1. www.VNMATH.com THI TUYN SINH LP 10CA CC TNH THNH PH NM HC 2012 2013MN TON thi vo lp 10 mn Ton nm 20121

2. www.VNMATH.comS GIO DC V O TOK THI TUYN SINH LP 10 THPT TP.HCM Nm hoc: 2012 2013 CHNH THC MN: TONThi gian lm bi: 120 phtBai 1: (2 im) Gii cc phng trnh v h phng trnh sau:a) 2 x 2 x 3 = 02 x 3 y = 7b) 3 x + 2 y = 4 c) x 4 + x 2 12 = 0 d) x 2 2 2 x 7 = 0Bai 2: (1,5 im) 1 21a) V th (P) ca hm s y =x v ng thng (D): y = x + 2 trn cng mt h trc to . 42 b) Tm to cc giao im ca (P) v (D) cu trn bng php tnh.Bai 3: (1,5 im) Thu gn cc biu thc sau: 1 2 x1 A=+ vi x > 0; x 1x + x x 1 x x B = (2 3) 26 + 15 3 (2 + 3) 26 15 3Bai 4: (1,5 im) Cho phng trnh x 2 2mx + m 2 = 0 (x l n s) a) Chng minh rng phng trnh lun lun c 2 nghim phn bit vi mi m. b) Gi x1, x2 l cc nghim ca phng trnh. 24 Tm m biu thc M = 2t gi tr nh nht x1 + x2 6 x1 x22Bai 5: (3,5 im) Cho ng trn (O) c tm O v im M nm ngoi ng trn (O). ng thng MO ct (O) ti E v F (ME 0; x 1B = (2 3) 26 + 15 3 (2 + 3) 26 15 3 thi vo lp 10 mn Ton nm 2012 3 4. www.VNMATH.com 1 1 =(2 3) 52 + 30 3 (2 + 3) 52 30 32 2 1 1 =(2 3) (3 3 + 5) 2 (2 + 3) (3 3 5) 22 2 11 =(2 3)(3 3 + 5) (2 + 3)(3 3 5) = 222Cu 4:a/ Phng trnh (1) c = m2 - 4m +8 = (m - 2)2 +4 > 0 vi mi m nn phng trnh (1) c 2 nghim phn bitvi mi m. bcb/ Do , theo Viet, vi mi m, ta c: S = = 2m ; P = = m 2 aa24 24 6M= = = ( x1 + x2 ) 2 8 x1 x2 4m 2 8m + 16 m 2 2m + 46= . Khi m = 1 ta c (m 1) 2 + 3 nh nht(m 1) 2 + 3 66 M = ln nht khi m = 1 M =nh nht khi m = 1 ( m 1) + 3 2 (m 1) 2 + 3K Vy M t gi tr nh nht l - 2 khi m = 1TCu 5Ba) V ta c do hai tam gic ng dng MAE v MBF Q MA MFASNn = MA.MB = ME.MF ME MB (Phng tch ca M i vi ng trn tm O)Vb) Do h thc lng trong ng trn ta c H2 MA.MB = MC , mt khc h thc lngM EOFtrong tam gic vung MCO ta cMH.MO = MC2 MA.MB = MH.MO Pnn t gic AHOB ni tip trong ng trn.c) Xt t gic MKSC ni tip trong ng Ctrn ng knh MS (c hai gc K v C vung).Vy ta c : MK2 = ME.MF = MC2 nn MK = MC. Do MF chnh l ng trung trc ca KC nn MS vung gc vi KC ti V.d) Do h thc lng trong ng trn ta c MA.MB = MV.MS ca ng trn tm Q.Tng t vi ng trn tm P ta cng c MV.MS = ME.MF nn PQ vung gc vi MS v l ng trung trcca VS (ng ni hai tm ca hai ng trn). Nn PQ cng i qua trung im ca KS (do nh l trung bnhca tam gic SKV). Vy 3 im T, Q, P thng hng.www.VNMATH.com thi vo lp 10 mn Ton nm 2012 4 5. www.VNMATH.comS GIO DC V O TOK THI TUYN SINH LP 10 THPT TP. NNG Nm hoc:2012 2013 MN: TON CHNH THCThi gian lm bi: 120 phtBi 1: (2,0 im)1) Gii phng trnh:(x + 1)(x + 2) = 0 2 x + y = 12) Gii h phng trnh: x 2 y = 7Bi 2: (1,0 im)Rt gn biu thc A = ( 10 2) 3 + 5yBi 3: (1,5 im) 2y=ax2Bit rng ng cong trong hnh v bn l mt parabol y = ax .1) Tm h s a.2) Gi M v N l cc giao im ca ng thng 2y = x + 4 vi parabol. Tm ta ca cc im M v N.Bi 4: (2,0 im)0 1 2 x22Cho phng trnh x 2x 3m = 0, vi m l tham s.1) Gii phng trnh khi m = 1.2) Tm tt c cc gi tr ca m phng trnh c hai nghim x1, x2 khc 0 v tha iu kin x1 x2 8 = . x2 x1 3Bi 5: (3,5 im)Cho hai ng trn (O) v (O) tip xc ngoi ti A. K tip tuyn chung ngoi BC, B (O), C (O).ng thng BO ct (O) ti im th hai l D.1) Ch`ng minh rng t gic COOB l mt hnh thang vung.2) Chng minh rng ba im A, C, D thng hng.3) T D k tip tuyn DE vi ng trn (O) (E l tip im). Chng minh rng DB = DE.BI GIIBi 1:1) (x + 1)(x + 2) = 0 x + 1 = 0 hay x + 2 = 0 x = -1 hay x = -2 2 x + y = 1 (1)5y = 15 ((1) 2(2)) y = 32) x 2 y = 7 (2) x = 7 + 2y x = 1Bi 2: A = ( 10 2) 3 + 5 = ( 5 1) 6 + 2 5 = ( 5 1) ( 5 + 1) 2 = ( 5 1)( 5 + 1) = 4Bi 3:1) Theo th ta c y(2) = 2 2 = a.22 a = 1 22) Phng trnh honh giao im ca y = x v ng thng y = x + 4 l :21 2 x + 4 = x x2 2x 8 = 0 x = -2 hay x = 42 y(-2) = 2 ; y(4) = 8. Vy ta cc im M v N l (-2 ; 2) v (4 ; 8).Bi 4:1) Khi m = 1, phng trnh thnh : x2 2x 3 = 0 x = -1 hay x = 3 (c dng ab + c = 0) thi vo lp 10 mn Ton nm 2012 5 6. www.VNMATH.comx1 x2 82) Vi x1, x2 0, ta c : = 3( x12 x2 ) = 8 x1 x2 3(x1 + x2)(x1 x2) = 8x1x2 2x2 x1 3 Ta c : a.c = -3m 0 nn 0, m 2bc Khi 0 ta c : x1 + x2 = = 2 v x1.x2 = = 3m 0 2aa iu kin phng trnh c 2 nghim 0 m m 0 > 0 v x1.x2 < 0 x1 < x2 Vi a = 1 x1 = b v x2 = b + x1 x2 = 2 = 2 1 + 3m 2 Do , ycbt 3(2)(2 1 + 3m 2 ) = 8(3m 2 ) v m 0 1 + 3m 2 = 2m 2 (hin nhin m = 0 khng l nghim) 4m4 3m2 1 = 0 m2 = 1 hay m2 = -1/4 (loi) m = 1Bi 5:BC OAOE D1) Theo tnh cht ca tip tuyn ta c OB, OC vung gc vi BC t gic COOB l hnh thang vung.2) Ta c gc ABC = gc BDC gc ABC + gc BCA = 900 gc BAC = 900 Mt khc, ta c gc BAD = 900 (ni tip na ng trn) Vy ta c gc DAC = 1800 nn 3 im D, A, C thng hng.3) Theo h thc lng trong tam gic vung DBC ta c DB2 = DA.DC Mt khc, theo h thc lng trong ng trn (chng minh bng tam gic ng dng) ta c DE 2 = DA.DC DB = DE.www.VNMATH.com thi vo lp 10 mn Ton nm 2012 6 7. www.VNMATH.comS GD&T K THI TUYN SINH LP 10 THPT NM HC 2012-2013VNH PHC THI MN : TON Thi gian lm bi 120 pht (khng k thi gian giao ) CHNH THCNgy thi: 21 thng 6 nm 2012 x36x 4Cu 1 (2,0 im). Cho biu thc :P=+ 2 x 1 x +1 x 1 1. Tm iu kin xc nh ca biu thc P. 2. Rt gn P 2 x + ay = 4Cu 2 (2,0 im). Cho h phng trnh : ax 3 y = 5 1. Gii h phng trnh vi a=1 2. Tm a h phng trnh c nghim duy nht.Cu 3 (2,0 im). Mt hnh ch nht c chiu rng bng mt na chiu di. Bit rng nu gim mi chiu i2m th din tch hnh ch nht cho gim i mt na. Tnh chiu di hnh ch nht cho.Cu 4 (3,0 im). Cho ng trn (O;R) (im O c nh, gi tr R khng i) v im M nm bn ngoi (O).K hai tip tuyn MB, MC (B,C l cc tip im ) ca (O) v tia Mx nm gia hai tia MO v MC. Qua B kng thng song song vi Mx, ng thng ny ct (O) ti im th hai l A. V ng knh BB ca (O).Qua O k ng thng vung gc vi BB,ng thng ny ct MC v BC ln lt ti K v E. Chng minhrng: 1. 4 im M,B,O,C cng nm trn mt ng trn. 2. on thng ME = R. 3. Khi im M di ng m OM = 2R th im K di ng trn mt ng trn c nh, ch r tm v bnknh ca ng trn .Cu 5 (1,0 im). Cho a,b,c l cc s dng tha mn a+ b + c =4. Chng minh rng : 4 a 3 + 4 b3 + 4 c3 > 2 2 S GD&T VNH PHCK THI TUYN SINH LP 10 THPT NM HC 2012-2013 P N THI MN : TON Ngy thi: 21 thng 6 nm 2012Cu p n, gi imC1.1(0,75 x 1 00,5 im) Biu thc P xc nh x + 1 0 x2 1 0 0,25 x 1 x 1 thi vo lp 10 mn Ton nm 2012 7 8. www.VNMATH.comC1.2 x3 6x 4x( x + 1) + 3( x 1) (6 x 4) 0,25P=+ =(1,25x 1 x + 1 ( x + 1)( x 1)( x + 1)( x 1)im)x + x + 3x 3 6 x + 4 2x 2 x +1 20,5 ==( x + 1)( x 1) ( x + 1)( x 1)0,5 ( x 1) 2x 1 = = (voi x 1) ( x + 1)( x 1) x + 1 2x + y = 4C2.10,25(1,0im) Vi a = 1, h phng trnh c dng: x 3y = 50,25 6x+ 3y = 12 7x = 70,25 0,25 x 3 y = 5 x 3y = 5 x = 1 x = 1 1 3y = 5 y = 2x= 1Vy vi a = 1, h phng trnh c nghim duy nht l: y= 2C2.20,25(1,0 x = 2im) 2x = 4 -Nu a = 0, h c dng: 5 => c nghim duy nht 0,25 3y = 5 y = 0,25 30,252a-Nu a 0 , h c nghim duy nht khi v ch khi: a 3 a 2 6 (lun ng, v a 2 0 vi mi a)Do , vi a 0 , h lun c nghim duy nht.Vy h phng trnh cho c nghim duy nht vi mi a.C3 (2,0Gi chiu di ca hnh ch nht cho l x (m), vi x > 4.0,25im) xV chiu rng bng na chiu di nn chiu rng l: (m)20,25 thi vo lp 10 mn Ton nm 20128 9. www.VNMATH.com x x2=> din tch hnh ch nht cho l: x. = (m2) 2 2Nu gim mi chiu i 2 m th chiu di, chiu rng ca hnh ch nht ln lt 0,25xl: x 2 va 2 (m)2khi , din tch hnh ch nht gim i mt na nn ta c phng trnh:x 1 x2 0,25( x 2)( 2) = 0,252 2 22xx2 2x x + 4 = x 2 12 x + 16 = 024 0,5.=> x1 = 6 + 2 5 (tho mn x>4);x 2 = 6 2 5 (loi v khng tho mn x>4) 0,25Vy chiu di ca hnh ch nht cho l 6 +2 5 (m).C4.11) Chng minh M, B, O, C cng thuc 1 ng trn B(1,0Ta c: MOB = 90 (v MB l tip tuyn)0im) MCO = 90 0 (v MC l tip tuyn)0,251O=> MBO + MCO = M 21= 900 + 900 = 1800K0,25=> T gic MBOC ni tip10,25E(v c tng 2 gc i =1800)BC=>4 im M, B, O, C cng thuc 1 ng trn0,25C4.22) Chng minh ME = R:(1,0Ta c MB//EO (v cng vung gc vi BB)im) => O1 = M1 (so le trong)M M1 = M2 (tnh cht 2 tip tuyn ct nhau) => M2 = O1 (1)0,25C/m c MO//EB (v cng vung gc vi BC)=> O1 = E1 (so le trong) (2) 0,25T (1), (2) => M2 = E1 => MOCE ni tip=> MEO = MCO = 900 0,25=> MEO = MBO = BOE = 900 => MBOE l hnh ch nht=> ME = OB = R (iu phi chng minh)0,25C4.33) Chng minh khi OM=2R th K di ng trn 1 ng trn c nh:(1,0Chng minh c Tam gic MBC u => BMC = 600im) => BOC = 12000,25=> KOC = 600 - O1 = 600 - M1 = 600 300 = 300 0,25Trong tam gic KOC vung ti C, ta c:OC OC 3 2 3RCosKOC = OK = 0= R: = 0,25OKCos3023M O c nh, R khng i => K di ng trn ng trn tm O, bn knh =2 3R 0,25 (iu phi chng minh)3C5 (1,0 44a 3 + 4 4b3 + 4 4c 3im)0,25= 4( a + b + c ) a 3 + 4 ( a + b + c ) b3 + 4 ( a + b + c ) c 3 0,25> 4 a 4 + 4 b4 + 4 c 4= a+b+c0,25=4 0,25 thi vo lp 10 mn Ton nm 2012 9 10. www.VNMATH.com44Do , a + b + c > 4 ==2 24 34 3 4 3 42Ch : -Cu 4, tha gi thit tia Mx v im A gy ri.-Mi cu u c cc cch lm khccu 5Cach 2: t x = 4 a; y = 4 b;z = 4 c => x, y , z > 0 v x4 + y4 + z4 = 4.BT cn CM tng ng: x3 + y3 + z3 > 2 2 hay 2 (x3 + y3 + z3 ) > 4 = x4 + y4 + z4 x3( 2 -x) + y3( 2 -y)+ z3( 2 -z) > 0 (*). Ta xt 2 trng hp:- Nu trong 3 s x, y, z tn ti it nht mt s 2 , gi s x 2 th x3 2 2 . Khi o: x3 + y3 + z3 > 2 2 ( do y, z > 0).- Nu c 3 s x, y, z u nh < 2 th BT(*) lun ung.Vy x + y3 + z3 > 2 2 c CM.3Cach 3: C th dng BT thc Csi kt hp phng php lm tri v nh gi cng cho kt qu nhng hidi, phc tp). thi vo lp 10 mn Ton nm 201210 11. www.VNMATH.comS GD V O TO K THI TUYN SINH VO 10 THPT NM HC 2012-2013 KLKMN THI : TONThi gian lm bi: 120 pht,(khng k giao ) CHNH THC Ngy thi: 22/06/2012Cu 1. (2,5)1) Gii phng trnh:a) 2x2 7x + 3 = 0.b) 9x4 + 5x2 4 = 0.2) Tm hm s y = ax + b, bit th hm s ca n i qua 2 im A(2;5) ; B(-2;-3).Cu 2. (1,5)1) Hai t i t A n B di 200km. Bit vn tc xe th nht nhanh hn vn tc xe th hai l 10km/h nnxe th nht n B sm hn xe th hai 1 gi. Tnh vn tc mi xe.2) Rt gn biu thc: A= 1 (1 ) x + x ; vi x 0. x + 1Cu 3. (1,5 )Cho phng trnh: x2 2(m+2)x + m2 + 4m +3 = 0.1) Chng minh rng : Phng trnh trn lun c hai nghim phn bit x1, x2 vi mi gi tr ca m.2) Tm gi tr ca m biu thc A = x1 + x 2 t gi tr nh nht. 22Cu 4. (3,5)Cho tam gic ABC c ba gc nhn ni tip ng trn tm O (AB < AC). Hai tip tuyn ti B v C ct nhau tiM. AM ct ng trn (O) ti im th hai D. E l trung im on AD. EC ct ng trn (O) ti im thhai F. Chng minh rng:1) T gic OEBM ni tip.2) MB2 = MA.MD. 3) BFC = MOC .4) BF // AMCu 5. (1)1 2Cho hai s dng x, y tha mn: x + 2y = 3. Chng minh rng: + 3x yBi gii s lc:Cu 1. (2,5)1) Gii phng trnh:a) 2x2 7x + 3 = 0. = (-7)2 4.2.3 = 25 > 07+ 5 x1 = = 3.4 = 5. Phng trnh c hai nghim phn bit:7 5 1 x2 = =4 2b) 9x4 + 5x2 4 = 0. t x2 = t , k : t 0.Ta c pt: 9t2 + 5t 4 = 0.a b + c = 0 t1 = - 1 (khng TMK, loi) 4 t2 = (TMK) 944 t2 = x2 = x = 4 = 2 .99 9 3 thi vo lp 10 mn Ton nm 2012 11 12. www.VNMATH.com 2 Vy phng trnh cho c hai nghim: x1,2 = 3 2a + b = 5 a = 22) th hm s y = ax + b i qua hai im A(2;5) v B(-2;-3) 2a + b = 3 b = 1Vy hm s cn tm l : y = 2x + 1Cu 2.1) Gi vn tc xe th hai l x (km/h). k: x > 0Vn tc xe th nht l x + 10 (km/h) 200Thi gian xe th nht i qung ng t A n B l : (gi)x + 10 200Thi gian xe th hai i qung ng t A n B l : (gi)x 200 200Xe th nht n B sm 1 gi so vi xe th hai nn ta c phng trnh: =1xx + 10Gii phng trnh ta c x1 = 40 , x2 = -50 ( loi)x1 = 40 (TMK). Vy vn tc xe th nht l 50km/h, vn tc xe th hai l 40km/h. x + 1 1 2) Rt gn biu thc: A = 1 1 ( x + 1 ) x+ x = x +1 ( x+ x )x = x + 1 () x x + 1 = x, vi x 0.Cu 3. (1,5 )Cho phng trnh: x2 2(m+2)x + m2 + 4m +3 = 0.1) Chng minh rng : Phng trnh trn lun c hai nghim phn bit x1, x2 vi mi gi tr ca m. 2 Ta c = (m + 2) m2 4m 3 = 1 > 0 vi mi m. Vy phng trnh cho lun c hai nghim phn bit x1, x2 vi mi gi tr ca m. 2) phng trnh cho lun c hai nghim phn bit x1, x2 vi mi gi tr ca m. Theo h thc Vi-t ta c : x1 + x 2 = 2(m + 2) x1.x 2 = m + 4m + 3 2 A = x1 + x 2 = (x1 + x2)2 2 x1x2 = 4(m + 2)2 2(m2 + 4m +3) = 2m2 + 8m+ 1022 = 2(m2 + 4m) + 10 = 2(m + 2)2 + 2 2 vi mi m. Suy ra minA = 2 m + 2 = 0 m = - 2Vy vi m = - 2 th A t min = 2 ACu 4. 1) Ta c EA = ED (gt) OE AD ( Quan h gia ng knh v dy)O C OEM = 900; OBM = 900 (Tnh cht tip tuyn) FE E v B cng nhn OM di mt gc vung T gic OEBM ni tip. 1B 2) Ta c MBD = s BD ( gc ni tip chn cung BD)2 D1MAB = s BD ( gc to bi tia tip tuyn v dy cung chn cung BD) 2 MBD = MAB . Xt tam gic MBD v tam gic MAB c:M thi vo lp 10 mn Ton nm 2012 12 13. www.VNMATH.com MB MD Gc M chung, MBD = MAB MBD ng dng vi MAB =MA MB MB2 = MA.MD 1 1 1 3) Ta c: MOC = BOC = s BC ( Tnh cht hai tip tuyn ct nhau); BFC = s BC (gc ni tip) 22 2 BFC = MOC .$ 4) T gic MFOC ni tip ( F + C = 1800) MFC = MOC ( hai gc ni tip cng chn cung MC), mt khc MOC = BFC (theo cu 3) BFC = MFC BF // AM. a 2 b2 ( a + b )2Cu 5. + xyx+ yTa c x + 2y = 3 x = 3 2y , v x dng nn 3 2y > 01 2 1 2y + 6 4y 3y(3 2y) 6(y 1)2Xt hiu+ 3= + 3= = 0 ( v y > 0 v 3 2y > 0)x y3 2y y y(3 2y) y(3 2y) x > 0,y > 0 x > 0,y > 0 1 1x = 1 + 3 du = xy ra x = 3 2y x = 1 x 2y y 1 = 0 y = 1 y = 1 thi vo lp 10 mn Ton nm 2012 13 14. www.VNMATH.comS GIO DC V O TOK THI TUYN SINH LP 10 THPT CHUYNHI DNG NGUYN TRI NM HC 2012- 2013 Mn thi: TON (khng chuyn) CHNH THC Thi gian lm bi: 120 phtNgy thi 19 thng 6 nm 2012 thi gm : 01 trangCu I (2,0 im) x 1 1) Gii phng trnh = x +1 . 3x 3 3 3 = 0 2) Gii h phng trnh .3 x + 2 y = 11Cu II ( 1,0 im) 1 1 a +1Rt gn biu thc P = + : vi a > 0 v a 4 .2 a -a2- a a-2 aCu III (1,0 im) Mt tam gic vung c chu vi l 30 cm, di hai cnh gc vung hn km nhau 7cm. Tnh di cccnh ca tam gic vung .Cu IV (2,0 im)1 2Trong mt phng ta Oxy, cho ng thng (d): y = 2x - m +1 v parabol (P): y = x .2 1) Tm m ng thng (d) i qua im A(-1; 3). 2) Tm m (d) ct (P) ti hai im phn bit c ta (x1; y1) v (x2; y2) sao cho x1x 2 ( y1 + y 2 ) + 48 = 0 .Cu V (3,0 im)Cho ng trn tm O ng knh AB. Trn ng trn ly im C sao cho AC < BC (C A). Cc tiptuyn ti B v C ca (O) ct nhau im D, AD ct (O) ti E (E A) .1) Chng minh BE2 = AE.DE.2) Qua C k ng thng song song vi BD ct AB ti H, DO ct BC ti F. Chng minh t gic CHOFni tip .3) Gi I l giao im ca AD v CH. Chng minh I l trung im ca CH.Cu VI ( 1,0 im) 1 1Cho 2 s dng a, b tha mn + = 2 . Tm gi tr ln nht ca biu thc a b 11Q= 4+ 4 .a + b + 2ab b + a + 2ba 22 22 thi vo lp 10 mn Ton nm 2012 14 15. www.VNMATH.comS GIO DC V O TO K THI TUYN SINH LP 10 THPT CHUYN NGUYN TRI HI DNG NM HC 2012 - 2013 HNG DN V BIU IM CHM MN TON (khng chuyn)Hng dn chm gm : 02 trangI) HNG DN CHUNG.- Th sinh lm bi theo cch ring nhng p ng c yu cu c bn vn cho im.- Vic chi tit im s (nu c) so vi biu im phi c thng nht trong Hi ng chm.- Sau khi cng im ton bi, im l n 0,25 im.II) P N V BIU IM CHM. CuNi dungim Cu I (2,0) 1) 1,0 im x 1 0,25 = x + 1 x 1 = 3( x + 1) 3 x 1 = 3x + 3 0,25 2x = 4 0,25 x = 2 .Vy phng trnh cho c mt nghim x = -2 0,25 2) 1,0 im x 3 3 3 = 0 (1) 0,25 T (1)=> x 3 = 3 3 3 x + 2 y = 11 (2) x=3 0,25 Thay x=3 vo (2)=> 3.3 + 2 y = 11 2y=20,25 y=1 . Vy h phng trnh cho c nghim (x;y)=(3;1)0,25 Cu II (1,0) 0,2511 a +1 P= + : ( a 2- a 2- a a 2 a )1+ a a2 a0,25 = a (2 a )a +1 = a ( a 2)0,25 a 2- a( ) a 2 0,25 = =-12- a Cu III Gi di cnh gc vung nh l x (cm) (iu kin 0< x < 15)0,25 (1,0)=> di cnh gc vung cn li l (x + 7 )(cm) V chu vi ca tam gic l 30cm nn di cnh huyn l 30(x + x +7)= 232x (cm) Theo nh l Py ta- go ta c phng trnhx 2 + (x + 7) 2 = (23 - 2x) 20,25 x 2 - 53x + 240 = 0 (1) Gii phng trnh (1) c nghim x = 5; x = 480,25 i chiu vi iu kin c x = 5 (TM k); x = 48 (khng TM k) 0,25 Vy di mt cnh gc vung l 5cm, di cnh gc vung cn li l 12 cm, di cnh huyn l 30 (5 + 12) = 13cm Cu IV (2,0)1) 1,0 imV (d) i qua im A(-1; 3) nn thay x = -1 v y = 3 vo hm s y = 2x m + 1 0,25 ta c 2.(-1) m +1 = 3 -1 m = 30,25 m = -40,25 thi vo lp 10 mn Ton nm 201215 16. www.VNMATH.comVy m = -4 th (d) i qua im A(-1; 3) 0,252) 1,0 im1 20,25Honh giao im ca (d) v (P) l nghim ca phng trnh x = 2 x m + 1 2 x 4 x + 2m 2 = 0 (1) ; (d) ct (P) ti hai im phn bit nn (1) c hai 20,25nghim phn bit > 0 6 2m > 0 m < 3V (x1; y1) v (x2; y2) l ta giao im ca (d) v (P) nn x1; x2 l nghim ca 0,25phng trnh (1) v y1 = 2 x1 m + 1 , y 2 = 2 x2 m + 1Theo h thc Vi-et ta c x1 + x 2 = 4, x1x 2 = 2m-2 .Thay y1,y2 vox1x 2 ( y1 +y 2 ) + 48 = 0 c x1x 2 ( 2x1 +2x 2 -2m+2 ) + 48 = 0 (2m - 2)(10 - 2m) + 48 = 0 m 2 - 6m - 7 = 0 m=-1(tha mn m OD l ng trung trc ca on BC => OFC=900I(1)F A H OB C CH // BD (gt), m AB BD (v BD l tip tuyn ca (O)) 0,25 => CH AB => OHC=900(2) 0,25 T (1) v (2) ta c OFC + OHC = 1800 => t gic CHOF ni tip 0,253)1,0 im C CH //BD=> HCB=CBD (hai gc v tr so le trong) m0,25 BCD cn ti D => CBD = DCB nn CB l tia phn gic ca HCDAI CI 0,25do CA CB => CA l tia phn gic gc ngoi nh C ca ICD =AD CD(3) AI HI0,25Trong ABD c HI // BD =>=(4) AD BD thi vo lp 10 mn Ton nm 201216 17. www.VNMATH.com CI HI 0,25 T (3) v (4) => = m CD=BD CI=HI I l trung im ca CHCD BD Cu VIVi a > 0; b > 0 ta c: (a 2 b) 2 0 a 4 2a 2b + b 2 0 a 4 + b 2 2a 2b 0,25 (1,0)11 a 4 + b 2 + 2ab 2 2a 2b + 2ab 2 a 4 + b 2 + 2ab 2 2ab a + b (1)() 110,25 Tng t c(2) . T (1) v (2)b + a + 2a b 2ab ( a + b )4 2 21 Q ab ( a + b )1 1110,25 V + = 2 a + b = 2ab m a + b 2 ab ab 1 Q 2 .a b2(ab)211 0,25 Khi a = b = 1 th Q = . Vy gi tr ln nht ca biu thc l22 thi vo lp 10 mn Ton nm 2012 17 18. www.VNMATH.comS GIO DC V O TO THI TUYN SINH VO LP 10 THPT TUYN QUANGNm hc 2011 - 2012 CHNH THCMN THI: TON Thi gian: 120 pht (khng k thi gian giao )Cu 1 (3,0 im) a) Gii phng trnh:x2 6x +9 =0 4 x 3 y = 6 b) Gii h phng trnh: 3 y + 4 x =10 c) Gii phng trnh: x 2 6 x + 9 = x 2011Cu 2 (2,5 im) Mt ca n chy xui dng t A n B ri chy ngc dng t B n A ht tt c 4 gi. Tnh vn tc ca n khi nc yn lng, bit rng qung sng AB di 30 km v vn tc dng nc l 4 km/gi.Cu 3 (2,5 im) Trn ng trn (O) ly hai im M, N sao cho M, O, N khng thng hng. Hai tip tuyn ti M , N vi ng trn (O) ct nhau ti A. T O k ng vung gc vi OM ct AN ti S. T A k ng vung gc vi AM ct ON ti I. Chng minh: a) SO = SA b) Tam gic OIA cnCu 4 (2,0 im). a) Tm nghim nguyn ca phng trnh: x2 + 2y2 + 2xy + 3y 4 = 0 b) Cho tam gic ABC vung ti A. Gi I l giao im cc ng phn gic trong. Bit AB = 5 cm, IC = 6 cm. Tnh BC.Hng dn chm, biu imMN THI: TON CHUNGNi dungim Cu 1 (3,0 im) a) Gii phng trnh: x2 6x +9 =01,0 Bi gii: Ta c = (3) 2 9 = 00,56 Phng trnh c nghim: x = =30,5 2 thi vo lp 10 mn Ton nm 2012 18 19. www.VNMATH.com4 x 3 y =6 (1) b) Gii h phng trnh: 1,03 y +4 x =10(2) Bi gii: Cng (1) v (2) ta c: 4x - 3y + 3y + 4x = 16 8x = 16 x = 2 0,5 x = 22 Thay x = 2 vo (1): 4. 2 3y = 6 y = . Tp nghim: 20,53y = 3 1,0 c) Gii phng trnh:x 2 6 x + 9 = x 2011 (3)( x 3) 2 Bi gii: Ta cx2 6 x + 9 = = x 30,5 Mt khc:x 2 6 x + 9 0 x 2011 0 x 2011 x 3 = x 30,5 Vy: (3) x 3 = x 2011 3 = 2011 . Phng trnh v nghim Cu 2 (2,5 im ) 2,5 Bi gii: Gi vn tc ca ca n khi nc yn lng l x km/gi ( x > 4)0,5 Vn tc ca ca n khi xui dng l x +4 (km/gi), khi ngc dng l x - 4 (km/gi). Thi gian30 ca n xui dng t A n B lgi, i ngc dng 0,5 x+430t B n A l gi. x4 3030 Theo bi ra ta c phng trnh:+ =4 (4) 0,5x+4 x4 (4) 30( x 4) +30( x +4) = 4( x +4)( x 4) x 2 15 x 16 = 0 x =1 0,5 hoc x = 16. Nghim x = -1 0). p dng nh l Pi-ta-go vo cc tam gic vung ABC v ACE ta c: AC2 = BC2 AB2 = x2 52= x2 -25 EC2 = AC2 + AE2 = x2 -25 + (x 5)2 = 2x2 10x (12: 2 )2 = 2x2 10xO,5x2 - 5x 36 = 0 Gii phng trnh ta c nghim x = 9 tho mn. Vy BC = 9 (cm) thi vo lp 10 mn Ton nm 201220 21. www.VNMATH.com thi vo lp 10 mn Ton nm 201221 22. www.VNMATH.com S GIO DC V O TO K THI TUYN SINH LP 10 THPT H NINm hoc: 2012 2013 CHNH THCMn thi: Ton Ngy thi: 21 thng 6 nm 2012 Thi gian lm bi: 120 phtBi I (2,5 im)x +4 1) Cho biu thc A = . Tnh gi tr ca A khi x = 36x +2 x 4 x + 16 2) Rt gn biu thc B = x +4 +: (vi x 0; x 16 )x 4 x +2 3) Vi cc ca biu thc A v B ni trn, hy tm cc gi tr ca x nguyn gi tr ca biu thc B(A 1) l s nguynBi II (2,0 im). Gii bi ton sau bng cch lp phng trnh hoc h phng trnh:12 Hai ngi cng lm chung mt cng vic tronggi th xong. Nu mi ngi lm mt mnh th ngi 5th nht hon thnh cng vic trong t hn ngi th hai l 2 gi. Hi nu lm mt mnh th mi ngi phi lmtrong bao nhiu thi gian xong cng vic?Bi III (1,5 im)2 1x + y = 2 1) Gii h phng trnh: 6 2 =1x y 2) Cho phng trnh: x (4m 1)x + 3m2 2m = 0 (n x). Tm m phng trnh c hai nghim phn2bit x1, x2 tha mn iu kin : x1 + x 2 = 722Bi IV (3,5 im) Cho ng trn (O; R) c ng knh AB. Bn knh CO vung gc vi AB, M l mt im bt k trncung nh AC (M khc A, C); BM ct AC ti H. Gi K l hnh chiu ca H trn AB. 1) Chng minh CBKH l t gic ni tip. 2) Chng minh ACM = ACK 3) Trn an thng BM ly im E sao cho BE = AM. Chng minh tam gic ECM l tam gic vungcn ti C 4) Gi d l tip tuyn ca (O) ti im A; cho P l im nm trn d sao cho hai im P, C nm trongAP.MBcng mt na mt phng b AB v = R . Chng minh ng thng PB i qua trung im ca onMAthng HKBi V (0,5 im). Vi x, y l cc s dng tha mn iu kin x 2y , tm gi tr nh nht ca biu thc:x 2 + y2 M= xy thi vo lp 10 mn Ton nm 201222 23. www.VNMATH.com GI P NBai I: (2,5 im)36 + 4 10 51) Vi x = 36, ta c : A ===36 + 2 8 42) Vi x , x 16 ta c : x( x 4) 4( x + 4) x + 2 (x + 16)( x + 2)x +2 B= x 16 + x 16 x + 16 = (x 16)(x + 16) = x 16 x +2 x +4 x +2 223) Ta c: B( A 1) = . 1 =. = .x 16 x + 2 x 16 x + 2 x 16 B( A 1) nguyn, x nguyn th x 16 l c ca 2, m (2) = { 1; 2 }Ta c bng gi tr tng ng: x 16 1 122 x17 15 18 14Kt hp K x 0, x 16 , B( A 1) nguyn th x { 14; 15; 17; 18 }Bai II: (2,0 im) 12Gi thi gian ngi th nht hon thnh mt mnh xong cng vic l x (gi), K x >5Th thi gian ngi th hai lm mt mnh xong cng vic l x + 2 (gi) 1 1Mi gi ngi th nht lm c (cv), ngi th hai lm c (cv) x x+2 1212 5V c hai ngi cng lm xong cng vic tronggi nn mi gi c hai i lm c 1: = (cv)5 5 12Do ta c phng trnh 115 + = x x + 2 12 x+2+ x 5 =x( x + 2) 12 5x2 14x 24 = 0 = 49 + 120 = 169, , = 137 13 67 + 13 20=> x ==(loi) v x = = = 4 (TMK)5 555Vy ngi th nht lm xong cng vic trong 4 gi,ngi th hai lm xong cng vic trong 4+2 = 6 gi.2 1x + y = 2Bai III: (1,5 im) 1)Gii h: , (K: x, y 0 ).6 2 =1x y4 24 610 x + y = 4 x + x = 4+1 x = 5 x = 2 x = 2 H 2 1.(TMK)6 2 =12 1 + =22 1 + = 2 2 y + = 2 y = 1x yx yx y Vy h c nghim (x;y)=(2;1). thi vo lp 10 mn Ton nm 201223 24. www.VNMATH.com2) + Phng trnh cho c = (4m 1) 12m2 + 8m = 4m2 + 1 > 0, m2 Vy phng trnh c 2 nghim phn bit m x1 + x2 = 4m 1 + Theo L Vi t, ta c: . x1x2 = 3m 2m 2Khi : x1 + x2 = 7 ( x1 + x2 ) 2x1x2 = 72 22 (4m 1)2 2(3m2 2m) = 7 10m2 4m 6 = 0 5m2 2m 3 = 03 Ta thy tng cc h s: a + b + c = 0 => m = 1 hay m =.5 Tr li: Vy....CBai IV: (3,5 im) MHEA K BO 1) Ta c HCB = 900 ( do chn na ng trn k AB)HKB = 900 (do K l hnh chiu ca H trn AB)=> HCB + HKB = 1800 nn t gic CBKH ni tip trong ng trn ng knh HB. 2) Ta c ACM = ABM (do cng chn ca (O))AM v ACK = HCK = HBK (v cng chn HK .ca trn k HB)Vy ACM = ACK 3) V OC AB nn C l im chnh gia ca cung AB AC = BC v sd = sd BC = 900 AC Xt 2 tam gic MAC v EBC c MA= EB(gt), AC = CB(cmt) v MAC = MBC v cng chn cung MC ca (O)MAC v EBC (cgc) CM = CE tam gic MCE cn ti C (1) Ta li c CMB = 450 (v chn cung CB = 900 ). CEM = CMB = 450 (tnh cht tam gic MCE cn ti C) M CME + CEM + MCE = 1800 (Tnh cht tng ba gc trong tam gic) MCE = 900 (2)T (1), (2) tam gic MCE l tam gic vung cn ti C (pcm). thi vo lp 10 mn Ton nm 2012 24 25. www.VNMATH.comS CMHP ENAKBO4) Gi S l giao im ca BM v ng thng (d), N l giao im ca BP vi HK.Xt PAM v OBM : AP.MB AP OBTheo gi thit ta c=R =(v c R = OB). MA MA MBMt khc ta c PAM = ABM (v cng chn cung ca (O)) AM PAM OBM AP OB = = 1 PA = PM .(do OB = OM = R) (3)PM OM V AMB = 900 (do chn na trn(O)) AMS = 900 M tam gic AMS vung ti M. PAM + PS = 900 v PMA + PMS = 900 M PMS = PS PS = PM (4)M PM = PA(cmt) nn PAM = PMAT (3) v (4) PA = PS hay P l trung im ca AS.NK BN HNNK HNV HK//AS (cng vung gc AB) nn theo L Ta-lt, ta c: = =hay = PA BP PS PA PSm PA = PS(cmt) NK = NH hay BP i qua trung im N ca HK. (pcm)Bai V: (0,5 im)Cch 1(khng s dng BT Co Si)x 2 + y 2 ( x 2 4 xy + 4 y 2 ) + 4 xy 3 y 2 ( x 2 y ) 2 + 4 xy 3 y 2 ( x 2 y ) 2 3yTa c M == ==+4xy xy xyxyxV (x 2y)2 0, du = xy ra x = 2y y 13 y 3x 2y , du = xy ra x = 2y x 2x 23 5T ta c M 0 + 4 - = , du = xy ra x = 2y2 2 thi vo lp 10 mn Ton nm 2012 25 26. www.VNMATH.com5Vy GTNN ca M l , t c khi x = 2y2Cch 2: x2 + y2 x2 y 2 x yx y 3xTa c M == += + = ( + )+xy xy xy y x4y x 4yx y x yx yV x, y > 0 , p dng bdt Co si cho 2 s dng ; ta c + 2 . =1, 4y x4y x 4y xdu = xy ra x = 2y x 3 x 6 3 V x 2y 2 . = , du = xy ra x = 2y y 4 y 4 23 5T ta c M 1 + = , du = xy ra x = 2y2 2 5Vy GTNN ca M l , t c khi x = 2y 2Cch 3: x2 + y2 x2 y 2 x y x 4 y 3yTa c M = = + = + = ( + )xyxy xy y x y x x x 4yx 4yx 4yV x, y > 0 , p dng bdt Co si cho 2 s dng ;ta c + 2 .=4, y x y x y xdu = xy ra x = 2y y 13 y 3 V x 2y , du = xy ra x = 2y x 2x23 5T ta c M 4- = , du = xy ra x = 2y2 25Vy GTNN ca M l , t c khi x = 2y2Cch 4:4x2x23x 2 x 2x2+ y2+ y2 + + y2 + y2Ta c M = x + y = 42 23x 2 3x = 44 = 4 += 4 +xy xyxy xy4 xyxy 4y x2 2x2x2 2V x, y > 0 , p dng bdt Co si cho 2 s dng ; y ta c+ y2 2 . y = xy , 4 4 4du = xy ra x = 2y x3 x 6 3 V x 2y 2 . = , du = xy ra x = 2y y4 y 4 2 xy 3 3 5T ta c M + = 1+ = , du = xy ra x = 2y xy 2 2 2 5Vy GTNN ca M l , t c khi x = 2y 2 thi vo lp 10 mn Ton nm 2012 26 27. www.VNMATH.com CHNH THC thi vo lp 10 mn Ton nm 201227 28. www.VNMATH.com thi vo lp 10 mn Ton nm 201228 29. www.VNMATH.com thi vo lp 10 mn Ton nm 201229 30. www.VNMATH.com thi vo lp 10 mn Ton nm 201230 31. www.VNMATH.com thi vo lp 10 mn Ton nm 201231 32. www.VNMATH.com thi vo lp 10 mn Ton nm 201232 33. www.VNMATH.com thi vo lp 10 mn Ton nm 201233 34. www.VNMATH.comS GIO DC O TOK THI VO LP 10 CHUYN LAM SN THANH HONM HC 2012 - 2013 CHNH THCMn thi : TON( gm c 01 trang)(Mn chung cho tt cc th sinh)Thi gian lm bi :120 pht (Khng k thi gian giao )Ngy thi : 17 thng 6 nm 2012Cu 1: (2.0 im ) Cho biu thc : a +1 a 1 1 P= a 1 +4 a 2a a , (Vi a > 0 , a 1) a +1 2 1. Chng minh rng : P = a 1 2. Tm gi tr ca a P = aCu 2 (2,0 im ) : Trong mt phng to Oxy, cho Parabol (P) : y = x2 v ng thng (d) : y = 2x + 3 1. Chng minh rng (d) v (P) c hai im chung phn bit 2. Gi A v B l cc im chung ca (d) v (P) . Tnh din tch tam gic OAB ( O l gc to )Cu 3 (2.0 im) : Cho phng trnh : x2 + 2mx + m2 2m + 4 = 0 1. Gii phng trnh khi m = 4 2. Tm m phng trnh c hai nghim phn bitCu 4 (3.0 im) : Cho ng trn (O) c ng knh AB c nh, M l mt im thuc (O) ( M khc A v B ) .Cc tip tuyn ca (O) ti A v M ct nhau C. ng trn (I) i qua M v tip xc vi ng thng AC ti C.CD l ng knh ca (I). Chng minh rng: 1. Ba im O, M, D thng hng 2. Tam gic COD l tam gic cn 3. ng thng i qua D v vung gc vi BC lun i qua mt im c nh khi M di ng trn ngtrn (O)Cu 5 (1.0 im) : Cho a,b,c l cc s dng khng m tho mn : a + b + c = 32 2 2abc 1+ 2+ 2Chng minh rng : a + 2b + 3 b + 2c + 3 c + 2a + 3 2 2 thi vo lp 10 mn Ton nm 201234 35. www.VNMATH.comBI GIICUNI DUNG IM 21. Chng minh rng : P = a 1 a +1 a 1 1P= a 1 a + 1 + 4 a 2a a () ()()( ). 2 2 a +1 a 1 + 4 aa +1a 1 1P=( a +1 )(a 1 )2a a1.0 a + 2 a + 1 a + 2 a 1 + 4a a 4 a 1P=.( a +1 )( ) a 1 2a a 1 4a a 12P= .=(PCM)a 1 2a a a 12. Tm gi tr ca a P = a. P = a2= a => a 2 a 2 = 0=> a 1 .Ta c 1 + 1 + (-2) = 0, nn phng trnh c 2 nghim 1.0a1 = -1 < 0 (khng tho mn iu kin) - Loic 2 = =2a2 = a 1 (Tho mn iu kin)Vy a = 2 th P = a 21. Chng minh rng (d) v (P) c hai im chung phn bitHonh giao im ng thng (d) v Parabol (P) l nghim ca phng trnhx2 = 2x + 3 => x2 2x 3 = 0 c a b + c = 0Nn phng trnh c hai nghim phn bit c 31.0 = =3x1 = -1 v x2 = a 1Vi x1 = -1 => y1 = (-1)2 = 1 => A (-1; 1)Vi x2 = 3 => y2 = 32 = 9 => B (3; 9)Vy (d) v (P) c hai im chung phn bit A v B2. Gi A v B l cc im chung ca (d) v (P) . Tnh din tch tam gic OAB ( O l1.0gc to )Ta biu din cc im A v B trn mt phng to Oxy nh hnh v9BA 1 D C -1 03 thi vo lp 10 mn Ton nm 201235 36. www.VNMATH.comAD + BC1+ 9S ABCD = .DC = .4 = 2022 BC.CO 9.3S BOC = == 13,52 2 AD.DO 1.1S AOD == = 0,52 2Theo cng thc cng din tch ta c:S(ABC) = S(ABCD) - S(BCO) - S(ADO)= 20 13,5 0,5 = 6 (vdt)1. Khi m = 4, ta c phng trnhx2 + 8x + 12 = 0 c = 16 12 = 4 > 01.0Vy phng trnh c hai nghim phn bitx1 = - 4 + 2 = - 2 v x2 = - 4 - 2 = - 6 32. Tm m phng trnh c hai nghim phn bitx2 + 2mx + m2 2m + 4 = 0C D = m2 (m2 2m + 4) = 2m 41.0 phng trnh c hai nghim phn bit th D > 0=> 2m 4 > 0 => 2(m 2) > 0 => m 2 > 0 => m > 2Vy vi m > 2 th phng trnh c hai nghim phn bit 4CID N H M 21.0AK 1B O1. Ba im O, M, D thng hng:Ta c MC l tip tuyn ca ng trn (O) MC MO (1) Xt ng trn (I) : Ta c CMD = 900 MC MD (2)T (1) v (2) => MO // MD MO v MD trng nhau O, M, D thng hng www.VNMATH.com2. Tam gic COD l tam gic cnCA l tip tuyn ca ng trn (O) CA AB(3)ng trn (I) tip xc vi AC ti C CA CD(4) T (3) v (4) CD // AB => DCO = COA (*) 1.0 ( Hai gc so le trong) CA, CM l hai tip tuyn ct nhau ca (O) COA = COD (**) T (*) v (**) DOC = DCO Tam gic COD cn ti D3. ng thng i qua D v vung gc vi BC lun i qua mt im c nh khi M di 1.0 thi vo lp 10 mn Ton nm 2012 36 37. www.VNMATH.comng trn ng trn (O)* Gi chn ng vung gc h t D ti BC l H. CHD = 900 H (I) (Bi tonqu tch)DH ko di ct AB ti K.Gi N l giao im ca CO v ng trn (I) CND = 900=> NC = NO COD can tai D Ta c t gic NHOK ni tip V c H 2 = O1 = DCO ( Cng b vi gc DHN) NHO + NKO = 1800 (5) * Ta c : NDH = NCH (Cng chn cung NH ca ng trn (I)) (CBO = HND = HCD DHN)COB (g.g)HN OB = HD OCOB OA HN ON ... = =M ONH = CDHOC OC HD CDOA CN ON ... ==OC CD CD NHO DHC (c.g.c) NHO = 90 M NHO + NKO = 1800 (5) NKO = 900 , NK AB NK // AC 0 K l trung im ca OA c nh (PCM) 5Cu 5 (1.0 im) : Cho a,b,c l cc s dng khng m tho mn : a + b + c = 3 2 221.0 ab c 1Chng minh rng : 2+ 2+ 2 a + 2b + 3 b + 2c + 3 c + 2a + 3 2a 2 b2 ( a + b )a 2 b2 c2 ( a + b + c )22* C/M b : + v + + . xyx+ yx y x x+ y+zTht vy a 2 b2 ( a + b ) 2 ( a 2 y + b 2 x ) ( x + y ) xy ( a + b ) ( ay bx ) 0 2 2 + x y x+ y(ng) PCM a 2 b2 c2 ( a + b + c ) 2p dng 2 ln , ta c:+ + x y x x+ y+z* Ta c : a + 2b + 3 = a + 2b + 1 + 2 2a + 2b + 2 , tng t Ta c: 2 2ab c abcA= 2 + 2+ 2+ +a + 2b + 3 b + 2c + 3 c + 2a + 3 2a + 2b + 2 2b + 2c + 2 2c + 2a + 2 1 a bc A ++ (1) 2 a + b +1 b + c +1 c + a +1 1444442444443Bab cTa chng minh++ 1 a + b +1 b + c +1 c + a +1 thi vo lp 10 mn Ton nm 2012 37 38. www.VNMATH.comab c1+1+ 1 2a + b +1 b + c +1c + a +1 b 1c 1 a 1+ + 2a + b +1 b + c +1 c + a +1b +1c +1a +1+ +2a + b +1 b + c +1 c + a +1 ( b + 1)( c + 1)( a + 1)22 2 + + 2(2)(1444444444 +24444444444 ) a + b + 1) ( b + 1) ( b + c 1) ( c + 1) ( c + a + 1) ( a + 143 3 B* p dng B trn ta c:( a + b + c + 3) 2 3 B ( a + b + 1) ( b + 1) + ( b + c + 1) ( c + 1) + ( c + a + 1) ( a + 1) ( a + b + c + 3) 2 3 B (3)a 2 + b 2 + c 2 + ab + bc + ca + 3(a + b + c ) + 3* M:2 a 2 + b 2 + c 2 + ab + bc + ca + 3(a + b + c) + 3 = 2a 2 + 2b 2 + 2c 2 + 2ab + 2bc + 2ca + 6a + 6b + 6c + 6 = 2a 2 + 2b 2 + 2c 2 + 2ab + 2bc + 2ca + 6a + 6b + 6c + 6 ( Do : a 2 + b 2 + c 2 = 3) = a 2 + b 2 + c 2 + 2ab + 2bc + 2ca + 6a + 6b + 6c + 9 = ( a + b + c + 3)2( a + b + c + 3) 2 =2 (4)a 2 + b 2 + c 2 + ab + bc + ca + 3(a + b + c) + 3T (3) v (4) (2)Kt hp (2) v (1) ta c iu phi chng minh.Du = xy ra khi a = b = c = 1 thi vo lp 10 mn Ton nm 201238 39. www.VNMATH.com S GIO DC V O TOK THI TUYN SINH LP 10 THPT THNH PH CN TH NM HC 2012-2013Kha ngy:21/6/2012MN: TON CHNH THC Thi gian lm bi: 120 pht (khng k thi gian pht )Cu 1: (2,0 im) Gii h phng trnh , cc phng trnh sau y: x + y = 43 1. 3 x 2 y = 19 2. x + 5 = 2 x 18 3. x 2 12 x + 36 = 0 4. x 2011 + 4 x 8044 = 3Cu 2: (1,5 im)1 1 a +1 Cho biu thc: K = 2 : 2 (vi a > 0, a 1 ) a 1a a a 1. Rt gn biu thc K. 2. Tm a K = 2012 .Cu 3: (1,5 im) Cho phng trnh (n s x): x 4 x m + 3 = 0 ( *) .2 2 1. Chng minh phng trnh (*) lun c hai nghim phn bit vi mi m. 2. Tm gi tr ca m phng trnh (*) c hai nghim x1 , x2 tha x2 = 5 x1 .Cu 4: (1,5 im) Mt t d nh i t A n B cch nhau 120 km trong mt thi gian quy nh. Sau khi i c 1 gi th t b chn bi xe cu ha 10 pht. Do n B ng hn xe phi tng vn tc thm 6 km/h. Tnh vn tc lc u ca t.Cu 5: (3,5 im) Cho ng trn ( O ) , t im A ngoi ng trn v hai tip tuyn AB v AC ( B, C l cc tip im). OA ct BC ti E. 1. Chng minh t gic ABOC ni tip. 2. Chng minh BC vung gc vi OA v BA.BE = AE.BO . 3. Gi I l trung im ca BE , ng thng qua I v vung gc OI ct cc tia AB, AC theo th t ti D v F . Chng minh IDO = BCO v DOF cn ti O . 4. Chng minh F l trung im ca AC . GI GII: www.VNMATH.comCu 1: (2,0 im) Gii h phng trnh , cc phng trnh sau y: x + y = 43 2 x + 2 y = 865 x = 105 x = 21 1. 3 x 2 y = 193 x 2 y = 19 x + y = 43 y = 22 thi vo lp 10 mn Ton nm 2012 39 40. www.VNMATH.com 2. x + 5 = 2 x 18 ; K : x 9 x = 23(TMK ) x + 5 = 2 x 18 x + 5 = 2 x + 18 x = 13 ( KTMK )3 3. x 12 x + 36 = 0 ( x 6) = 0 x = 6 22x 2011 + 4 x 8044 = 3; K : x 2011 4. 3 x 2011 = 3 x = 2012(TMK )Cu 2: (1,5 im)1 1 a +1 Cho biu thc: K = 2 : 2 (vi a > 0, a 1 ) a 1a a a 11 a +1 a a +1 a +1 K = 2 : a 2 a = 2 : a 1a a ( a 1) a (a 1) = 21: 1 a ( a 1) a( a 1) = 21 ( ) : a( a 1) = 2 aa ( a 1) K = 2012 2 a = 2012 a = 503 (TMK)Cu 3: (1,5 im) Cho phng trnh (n s x):.x 2 4 x m 2 + 3 = 0 ( *) 1. = 16 + 4m 2 12 = 4m 2 + 4 4 > 0; m Vy (*) lun co hai nghim phn bit vi mi m. 2. Tm gi tr ca m phng trnh (*) c hai nghim x1 , x2 tha x2 = 5 x1 . Theo h thc VI-ET co :x1.x2 = - m2 + 3 ;x1+ x2 = 4; ma x2 = 5 x1 => x1 = - 1 ; x2 = 5 Thay x1 = - 1 ; x2 = 5 vao x1.x2 = - m2 + 3 => m = 2 2Cu 4: (1,5 im) 120 Goi x (km/h) la vt d inh; x > 0 => Thi gian d inh :( h) xSau 1 h t i c x km => quang ng con lai 120 x ( km)Vt luc sau: x + 6 ( km/h) 1 120 x 120Pt 1 + +==> x = 48 (TMK) => KL 6x+6 xHD C3Tam giac BOC cn tai O => goc OBC = goc OCBT giac OIBD co goc OID = goc OBD = 900 nn OIBD ni tip => goc ODI = goc OBI Do o IDO = BCO Lai co FIOC ni tip ; nn goc IFO = goc ICOSuy ra goc OPF = goc OFP ; vy DOF cn ti O .HD C4Xet t giac BPFE co IB = IE ; IP = IF ( Tam giac OPF cn co OI la ng cao=> )Nn BPEF la Hinh binh hanh => BP // FETam giac ABC co EB = EC ; BA // FE; nn EF la TB cua tam giac ABC => FA = FC thi vo lp 10 mn Ton nm 201240 41. www.VNMATH.comS GD T NGH AN thi vo THPT nm hc 2012 - 2013Mn thi: Ton CHNH THCThi gian 120 phtNgy thi 24/ 06/ 2012Cu 1: 2,5 im: 1 1 x 2Cho biu thc A = + . x +2 x 2x a) Tm iu kin xc nh v t gn A. 1b) Tm tt c cc gi tr ca x A > 2 7c) Tm tt c cc gi tr ca x B = A t gi tr nguyn. 3Cu 2: 1,5 im:Qung ng AB di 156 km. Mt ngi i xe my t A, mt ngi i xe p t B. Hai xe xut pht cngmt lc v sau 3 gi gp nhau. Bit rng vn tc ca ngi I xe my nhanh hn vn tc ca ngi Ixe p l 28 km/h. Tnh vn tc ca mi xe?Cu 3: 2 im:Chjo phng trnh: x2 2(m-1)x + m2 6 =0 ( m l tham s). a) GiI phng trnh khi m = 3 b) Tm m phng trnh c hai nghim x1, x2 tha mn x1 + x2 = 1622Cu 4: 4 imCho im M nm ngoi ng trn tm O. V tip tuyn MA, MB vi ng trn (A, B l cc tipim). V ct tuyn MCD khng I qua tm O ( C nm gia M v D), OM ct AB v (O) ln lt ti Hv I. Chng minh. a) T gic MAOB ni tip. b) MC.MD = MA2 c) OH.OM + MC.MD = MO2 d) CI l tia phn gic gc MCH. thi vo lp 10 mn Ton nm 2012 41 42. www.VNMATH.com HNG DN GII www.VNMATH.comCu 1: (2,5 im)a, Vi x > 0 v x 4, ta c: 1 1 x 2 x 2+ x +2 x 22A= + .=.= ... = x +2 x 2x( x + 2)( x 2) xx +22 2 1b, A = > ... x > 4. x +2x +2 2 7214c, B = .= l mt s nguyn ... x + 2 l c ca 14 hay x + 2 = 1, x +2 = 3 x +2 3( x + 2) 7, x + 2 = 14.(Gii cc pt trn v tm x)Cu 2: (1,5 im)Gi vn tc ca xe p l x (km/h), iu kin x > 0Th vn tc ca xe my l x + 28 (km/h)Trong 3 gi: + Xe p i c qung ng 3x (km), + Xe my i c qung ng 3(x + 28) (km), theo bi ra ta c phng trnh:3x + 3(x + 28) = 156Gii tm x = 12 (TMK)Tr li: Vn tc ca xe p l 12 km/h v vn tc ca xe my l 12 + 28 = 40 (km/h)Cu 3: (2,0 im)a, Thay x = 3 vo phng trnh x2 - 2(m - 1)x + m2 - 6 = 0 v gii phng trnh: x2 - 4x + 3 = 0 bng nhiu cch v tm c nghim x1 = 1, x2 = 3.b, Theo h thc Vit, gi x1, x2 l hai nghim ca phng trnhx2 - 2(m - 1)x + m2 - 6 = 0 , ta c: thi vo lp 10 mn Ton nm 201242 43. www.VNMATH.com x1 + x2 = 2(m 1) x1.x2 = m 6 2 v x12 + x22 = (x1 + x2)2 - 2x1.x2 = 16 Thay vo gii v tm c m = 0, m = -4Cu 4: (4,0 im).T vit GT-KLADCM OI H H Ba, V MA, MB l cc tip tuyn ca ng trn (O) ti A v B nn cc gc ca t gic MAOB vung ti A vB, nn ni tip c ng trn. b, MAC v MDA c chung M v MAC = MDA (cng chn AC ), nn ng dng. T suy raMA MD= MC.MD = MA2 (fcm)MC MAc, MAO v AHO ng dng v c chung gc O v AMO = HAO (cng chn hai cung bng nhau cang trn ni tip t gic MAOB). Suy ra OH.OM = OA2p dng nh l Pitago vo tam gic vung MAO v cc h thc OH.OM = OA 2 MC.MD = MA2 suy ra iuphi chng minh.MH MCd, T MH.OM = MA2, MC.MD = MA2 suy ra MH.OM = MC.MD = (*)MD MO Trong MHC v MDO c (*) v DMO chung nn ng dng.MC MO MO MC MO == hay = (1)HC MD OA CH OA Ta li c MAI = IAH (cng chn hai cung bng nhau) AI l phn gic ca MAH .MI MATheo t/c ng phn gic ca tam gic, ta c: = (2)IH AH MHA v MAO c OMA chung v MHA = MAO = 900 do ng dng (g.g) thi vo lp 10 mn Ton nm 2012 43 44. www.VNMATH.comMO MA = (3)OA AHMC MIT (1), (2), (3) suy ra = suy ra CI l tia phn gic ca gc MCHCH IHS GIO DC V O TO K THI TUYN SINH LP 10 THPTH NAMNM HC 2012 2013 Mn: Ton CHNH THC Thi gian lm bi: 120 phtNgy thi : 22/06/2012Cu 1 (1,5 im) Rt gn cc biu thc sau: a) A = 2 5 + 3 45 500 8 2 12 b) B = 83 1Cu 2: (2 im)a) Gii phng trnh: x2 5x + 4 = 0 3x y = 1b) Gii h phng trnh: x + 2y = 5Cu 3: (2 im)Trong mt phng to Oxy cho Parabol (P) c phng trnh: y = x2 v ng thng (d) c phng trnh: y =2mx 2m + 3 (m l tham s)a) Tm to cc im thuc (P) bit tung ca chng bng 2b) Chng minh rng (P) v (d) ct nhau ti hai im phn bit vi mi m.Gi y1 , y 2 l cc tung giao im ca (P) v (d), tm m y1 + y 2 < 9Cu 4: (3,5 im)Cho ng trn tm O, ng knh AB. Trn tip tuyn ca ng trn (O) ti A ly im M ( M khc A). TM v tip tuyn th hai MC vi (O) (C l tip im). K CH vung gc vi AB ( H AB ), MB ct (O) ti imth hai l K v ct CH ti N. Chng minh rng:a) T gic AKNH l t gic ni tip.b) AM2 = MK.MBc) Gc KAC bng gc OMBd) N l trung im ca CH.Cu 5(1 im)Cho ba s thc a, b, c tho mn a 1; b 4;c 9Tm gi tr ln nht ca biu thc : bc a 1 + ca b 4 + ab c 9P= abc thi vo lp 10 mn Ton nm 201244 45. www.VNMATH.com thi vo lp 10 mn Ton nm 201245 46. www.VNMATH.com thi vo lp 10 mn Ton nm 201246 47. www.VNMATH.com S GIO DC V OTO K THI TUYN SINH LP 10 THPT NM HC 2012-2013 KHA NGY : 19/6/2012 QUNG TR MN : TON CHNH THC Thi gian lm bi: 120 pht (khng k thi gian giao )Cu 1:(2 im)1.Rt gn cc biu thc (khng dng my tnh cm tay): a) 2 50 - 18 1 1 1b) P = + a 1 , vi a 0,a 1 a 1a +1 2.Gii h phng trnh (khng dng my tnh cm tay): x+ y = 4 2x y = 5Cu 2:(1,5 im) Gi x1, x2 l hai nghim ca phng trnh x 2 5 x 3 = 0 .Khng gii phng trnh, tnh gi tr cc biuthc sau: 1a, x1 + x2 b, c, x12 + x 2 2x1 + x 2Cu 3:(1,5 im)Trn mt phng ta , gi (P) l th hm s y = x 2a, V (P)b, Tm ta giao im ca (P) v ng thng d: y = -2x+3Cu 4:(1,5 im)Hai xe khi hnh cng mt lc i t a im A n a im B cch nhau 100km. Xe th nht chynhanh hn xe th hai 10km/h nn n B sm hm 30 pht, Tnh vn tc mi xe.Cu 5:(3,5 im)Cho ng trn (O). ng thng (d) khng i qua tm (O) ct ng trn ti hai im A v B theo tht, C l im thuc (d) ngoi ng trn (O). V ng knh PQ vung gc vi dy AB ti D ( P thuc cungln AB), Tia CP ct ng trn (O) ti im th hai l I, AB ct IQ ti K.a) Chng minh t gic PDKI ni tip ng trn.b) Chng minh CI.CP = CK.CDc) Chng minh IC l phn gic ca gc ngoi nh I ca tam gic AIB.d) Cho ba im A, B, C c nh. ng trn (O) thay i nhng vn i qua A v B. Chng minhrng IQ lun i qua mt im c nh.e) www.VNMATH.com thi vo lp 10 mn Ton nm 2012 47 48. www.VNMATH.com S GIO DC O TOK THI TUYN SINH VO LP 10 THPT NINH THUN NM HC 2012 2013Kha ngy: 24 6 2012 CHNH THC Mn thi: TON Thi gian lm bi: 120 phtBi 1: (2,0 im)2 x + y = 3 a) Gii h phng trnh: x + 3y = 4 b) Xc nh cc gi tr ca m h phng trnh sau v nghim:(m + 2) x + (m + 1) y = 3 ( m l tham s)x + 3y = 4Bi 2: (3,0 im)Cho hai hm s y = x2 v y = x + 2. a) V th hai hm s cho trn cng mt h trc ta Oxy. b) Bng php tnh hy xc nh ta cc giao im A, B ca hai th trn (im A c honh m). c) Tnh din tch ca tam gic OAB (O l gc ta )Bi 3: (1,0 im) Tnh gi tr ca biu thc H = ( 10 2) 3 + 5Bi 4: (3,0 im)Cho ng trn tm O, ng knh AC = 2R. T mt im E trn on OA (E khng trng vi A vO). K dy BD vung gc vi AC. K ng knh DI ca ng trn (O). a) Chng minh rng: AB = CI. b) Chng minh rng: EA2 + EB2 + EC2 + ED2 = 4R2 2R c) Tnh din tch ca a gic ABICD theo R khi OE =3Bi 5: (1,0 im)Cho tam gic ABC v cc trung tuyn AM, BN, CP. Chng minh rng:3 (AB + BC + CA) < AM + BN + CP < AB + BC + CA4P N:Bi 1: (2,0 im) 2 x + y = 3 2 x + y = 3 5 y = 5x = 1a) Gii h phng trnh: x + 3y = 42 x + 6 y = 8x + 3y = 4 y =1b) H phng trnh v nghim khi: m + 2 m +1m + 2 m +1 3 1 = 3 3m + 6 = m + 15 = m=1 3 4 m +1 3 4m + 4 9 2 3 4Bi 2: (3,0 im)a) V (d) v (P) trn cng mt h trc ta . x -2 -1 0 1 2 2y = x (P) 41 0 1 4 thi vo lp 10 mn Ton nm 2012 48 49. www.VNMATH.com x-2 0y = x + 2(d) 0 264 B21 A -21 2 -10-5O5 10 -2 -4 -6b) Ta giao im ca (P) v (d) l nghim ca h phng trnh: y = x2 x2 = x + 2 x2 x 2 = 0 x = 1; x2 = 2 1y = x + 2 y = x + 2y = x + 2 y1 = 1; y2 = 4Ta cc giao im ca (d) v (P): A (-1;1) v B (2;4) 111c) SOAB = .(1+4).3 -.1.1 - .2.4 = 3 222Bi 3: (1,0 im)H = ( 10 2) 3 + 5 = ( 5 1)6+2 5 = (5 1 )()5 +1 = 5 1 = 4Bi 4: (3,0 im) a) Chng minh rng: AB = CI. BITa c: BD AC (gt)DBI = 90 ( gc ni tip chn na ng trn) BD BI 0Do : AC // BI = CI AB = CIAB A EC Ob) Chng minh rng: EA2 + EB2 + EC2 + ED2 = 4R2V BD AC = AB AD nn AB = AD DTa c: EA2 + EB2 + EC2 + ED2 = AB2 + CD2 = AD2 + CD2 = AC2 = (2R)2 = 4R2 2R c) Tnh din tch ca a gic ABICD theo R khi OE =3 11SABICD = SABD + SABIC = .DE.AC + .EB.(BI + AC) 22 2RR 2R 5R* OE = AE = v EC =+R=3333 thi vo lp 10 mn Ton nm 2012 49 50. www.VNMATH.comR 5R 25R R 5 R 5* DE2 = AE.EC = . = DE =. Do : EB =3 3 9 3 3R 4R* BI = AC 2AE = 2R 2. =3 3 1 R 51 R 5 4RR 5 16 R 8R 2 5Vy: SABICD = ..2R +.( + 2R) = . =(vdt) 2 32 33 63 9Bi 5: (1,0 im)A Cho tam gic ABC v cc trung tuyn AM, BN, CP. Chng minh rng: 3 (AB + BC + CA) < AM + BN + CP < AB + BC + CA N 4 PG 1 1 1 CGi G l trng tm ca ABC, ta c: GM = B AM; GN = BN; GP = CP M 333V AM, BN, CP cc trung tuyn, nn: M, N, P ln lt l trung im ca BC, AC, ABDo : MN, NP, MP l cc ng trung bnh ca ABC1 1 1Nn: MN = AB; NP = BC; MP = AC2 2 2p dng bt ng thc tam gic, ta c:1 1* AM < MN + AN hay AM MN hay BN + AM > AB (4) 33211 1Tng t: BN + CP > BC (5)33 21 1 1CP + AM > AC (6)3 3 2T (4), (5), (6) suy ra:1 11111111BN + AM + BN + CP + CP + AM > AB + BC+ AC3 33333222 21 (AM + BN + CP) > (AB + AC + BC) 32 3 (AB + BC + CA) < AM + BN + CP(**) 4 3T (*), (**) suy ra: (AB + BC + CA) < AM + BN + CP < AB + BC + CA 4 thi vo lp 10 mn Ton nm 2012 50 51. www.VNMATH.com thi vo lp 10 mn Ton nm 201251 52. www.VNMATH.com CHNH THC thi vo lp 10 mn Ton nm 201252 53. www.VNMATH.com thi vo lp 10 mn Ton nm 201253 54. www.VNMATH.com thi vo lp 10 mn Ton nm 201254 55. www.VNMATH.comS GIO DC V OTO K THI TUYN SINH LP 10 THPT NM HC 2012-2013 Kha ngy : 24/6/2012 THA THIN HUMn thi : TON CHNH THCThi gian lm bi: 120 pht (khng k thi gian giao )Bi 1:(2,0 im) 5+3 53+ 3a).Cho biu thc: C = 5+ 3 +1 ( )5 + 3 . Chng t C = 3b) Gii phng trnh : 3 x 2 x 2 4 = 0Bi 2:(2,0 im) Cho hm s y = x2 c th (P) v ng thng (d) i qua im M (1;2) c h s gc k 0.a/ Chng minh rng vi mi gi tr k 0. ng thng (d) lun ct (P) ti hai im phn bit A v B.b/ Gi xA v xB l honh ca hai im A v B.Chng minh rng x A + x B x A .xB 2 = 0Bi 3:(2,0 im)a/ Mt xe la i t ga A n ga B.Sau 1 gi 40 pht, mt xe la khc i t ga A n ga B vi vntc ln hn vn tc ca xe la th nht l 5 km/h.Hai xe la gp nhau ti mt ga cch ga B 300km.Tm vn tc ca mi xe, bit rng qung ng st t ga A n ga B di 645 km.2 ( x + y ) = 5 ( x y )b/ Gii h phng trnh : 20 20x + y + x y = 7Bi 4:(3,0 im)Cho na ng trn (O) ng knh BC.Ly im A trn tia i ca tia CB.K tip tuyn AF vi nang trn (O) ( F l tip im), tia AF ct tia tip tuyn Bx ca na ng trn (O) ti D ( tia tiptuyn Bx nm trong na mt phng b BC cha na ng trn (O)) .Gi H l giao im ca BF viDO ; K l giao im th hai ca DC vi na ng trn (O).a/ Chng minh rng : AO.AB=AF.AD.b/ Chng minh t gic KHOC ni tip.BD DM OAc/ K OM BC ( M thuc on thng AD).Chng minh =1DM AMBi 5:(1,0 im) Cho hnh ch nht OABC, COB = 300 .Gi CH l ng cao ca tam gic30 0COB, CH=20 cm.Khi hnh ch nht OABC quay mt vng quanh cnh OC cnh ta c mt hnh tr, khi tam gic OHC to thnh hnh (H).Tnh thtch ca phn hnh tr nm bn ngoi hnh (H).(Cho 3,1416 )K H 12 cm BC thi vo lp 10 mn Ton nm 201255 56. www.VNMATH.com thi vo lp 10 mn Ton nm 201256 57. www.VNMATH.com thi vo lp 10 mn Ton nm 201257 58. www.VNMATH.com thi vo lp 10 mn Ton nm 201258 59. www.VNMATH.com thi vo lp 10 mn Ton nm 201259 60. www.VNMATH.comS GIAO DUC VA AO TAOKY THI TUYN SINH PHU THOVAO LP 10 TRUNG HOC PH THNGNM HOC 2012-2013 CHNH THC Mn toan Thi gian lam bai: 120 phut, khng k thi gian giao thi co 01 trang -------------------------------------------Cu 1 (2)a) Giai phng trinh 2x 5 =1b) Giai bt phng trinh 3x 1 > 5Cu 2 (2) 3x + y = 3a) Giai h phng trinh 2x y = 7116b) Chng minh rng+ =3+ 2 3 2 7Cu 3 (2)Cho phng trinh x2 2(m 3)x 1 = 0a) Giai phng trinh khi m = 1b) Tim m phng trinh co nghim x1 ; x2 ma biu thcA = x12 x1x2 + x22 at gia tri nho nht? Tim gia tri nho nht o.Cu 4 (3) Cho tam giac ABC vung tai A. Ly B lam tm ve ng tron tm B ban kinh AB.Ly C lam tm ve ngtron tm C ban kinh AC, hai ng tron nay ct nhau tai im th 2 la D.Ve AM, AN ln lt la cac dy cungcua ng tron (B) va (C) sao cho AM vung goc vi AN va D nm gia M; N.a) CMR: ABC=DBCb) CMR: ABDC la t giac ni tip.c) CMR: ba im M, D, N thng hangd) Xac inh vi tri cua cac dy AM; AN cua ng tron (B) va (C) sao cho oan MN co dai ln nht. x 2 5 y 2 8 y = 3Cu 5 (1) Giai H PT (2 x + 4 y 1) 2 x y 1 = (4 x 2 y 3) x + 2 y---------------------------Ht-------------------------- GI GIICu 1 (2) a) Giai phng trinh 2x 5 = 1b) Giai bt phng trinh 3x 1 > 5ap an a) x = 3 ; b) x > 2 3x + y = 3Cu 2 (2) a) Giai h phng trinh 2x y = 7 thi vo lp 10 mn Ton nm 2012 60 61. www.VNMATH.com 11 6b) Chng minh rng += 3+ 2 3 2 7ap an a) x = 2 ; y = 33 2 +3+ 2 6b) VT = = =VP (pcm) 92 7Cu 3 (2) Cho phng trinh x2 2(m 3)x 1 = 0 c) Giai phng trinh khi m = 1 d) Tim m phng trinh co nghim x1 ; x2 ma biu thcA = x12 x1x2 + x22 at gia tri nho nht? Tim gia tri nho nht o.ap an a) x1 = 2 5 ; x2 = 2 + 5 e) Thy h s cua pt : a = 1 ; c = A 1 pt lun co 2 nghimTheo vi- et ta co x1 + x2 =2(m 3) ; x1x2 = 1Ma A=x12 x1x2 + x22 = (x1 + x2 )2 3x1x2 = 4(m 3)2 + 3 3 GTNN cua A = 3 m = 3Cu 4 (3)Hng dna) Co AB = DB; AC = DC; BC chung ABC = DBC (c-c-c)b) ABC = DBC goc BAC =BDC = 900 ABDC la t giac ni tipc) Co gocA1 = gocM1 ( ABM cn tai B)AgocA4 = gocN2 ( ACN cn tai C) 1 gocA1 = gocA4 ( cung phu A2;3 ) 2 3 4 gocA1 = gocM1 =gocA4= gocN2M1 2 B gocA2 = gocN1 ( cung chn cung AD cua (C) )1 2 CLai co A1+A2 + A3 = 900 => M1 + N1 + A3 = 900Ma AMN vung tai A => M1 + N1 + M2 = 9001 2 3 => A3 = M2 => A3 = D1 4CDN cn tai C => N1;2 = D4 D 1 2 D2;3 + D1 + D4 =D2;3 + D1 + N1;2 = D2;3 + M2 + N1 + N2N = 900 + M2 + N1 + M1 ( M1 = N2) = 900 + 900 = 1800 M; D; N thng hang.d) AMN ng dang ABC (g-g)Ta co NM2 = AN2 +AM2 NM ln nht thi AN ; AM ln nhtMa AM; AN ln nht khi AM; AN ln lt la ng kinh cua (B) va (C)Vy khi AM; AN ln lt la ng kinh cua (B) va (C) thi NM ln nht. x 2 5 y 2 8 y = 3Cu 5 (1): Giai H PT (2 x + 4 y 1) 2 x y 1 = (4 x 2 y 3) x + 2 yHng dn x 2 5 y 2 8 y = 3 (2 x + 4 y 1) 2 x y 1 = (4 x 2 y 3) x + 2 y thi vo lp 10 mn Ton nm 201261 62. www.VNMATH.com x 2 5 y 2 8 y = 3(1) (2 < x + 2 y > 1) 2 x y 1 = (2 < 2 x y 1 > 1) x + 2 y (2)T (2) t x +2y = a ; 2xy 1 = b (a:b 0)Ta dc (2a-1) b =(2b 1) a ( a b )(2 ab +1) = 0 a = b x = 3y + 1 thay vao (1) ta dc2y2 y 1= 0 => y1 = 1 ; y2 = 1/2=> x1 = 4 ; x2 = 1/2Thy x2 + 2y2 = 1 < 0 (loai)Vy h co nghim (x; y) = (4 ; 1) S gio dc v o to k thi tuyn sinh vo lp 10 thpt chuynHng yn Nm hc 2012 - 2013 Mn thi: Ton CHNH THC(Dnh cho th sinh d thi cc lp chuyn: Ton, Tin) ( thi c 01 trang)Thi gian lm bi: 150 phtBi 1: (2 im)a) Cho A = 20122 + 20122.20132 + 20132 . Chng minh A l mt s t nhin. 2 1 xx + y2 + y = 3b) Gii h phng trnh x + 1 + x = 3y yBi 2: (2 im)a) Cho Parbol (P): y = x2 v ng thng (d): y = (m +2)x m + 6. Tm m ng thng (d) ct Parabol (P) ti hai im phn bit c honh dng.b) Gii phng trnh: 5 + x + 2 (4 x)(2x 2) = 4( 4 x + 2x 2)Bi 3: (2 im)a) Tm tt c cc s hu t x sao cho A = x2 + x+ 6 l mt s chnh phng. (x 3 + y3 ) (x 2 + y 2 )b) Cho x > 1 v y > 1. Chng minh rng :8(x 1)(y 1)Bi 4 (3 im) thi vo lp 10 mn Ton nm 201262 63. www.VNMATH.comCho tam gic ABC nhn ni tip ng trn tm O, ng cao BE v CF. Tip tuyn ti B v C ct nhau ti S,gi BC v OS ct nhau ti Ma) Chng minh AB. MB = AE.BS b) Hai tam gic AEM v ABS ng dng c) Gi AM ct EF ti N, AS ct BC ti P. CMR NP vung gc vi BCBi 5: (1 im)Trong mt gii bng c 12 i tham d, thi u vng trn mt lt (hai i bt k thi u vi nhau ng mttrn).a) Chng minh rng sau 4 vng u (mi i thi u ng 4 trn) lun tm c ba i bng i mt cha thiu vi nhau.b) Khng nh trn cn ng khng nu cc i thi u 5 trn?HNG DN GIIBi 1: (2 im) a) Cho A = 20122 + 20122.20132 + 20132t 2012 = a, ta c20122 + 20122.20132 + 20132 = a + a (a + 1) + (a + 1)2 2 22= (a 2 + a + 1)2 = a 2 + a + 1x 2 1 x 2 1 xy = ax + y2 + y = 3 x + = 3 y y b) t Ta c x + 1 = bx + 1 + x = 3x + 1 + x = 3yy y y y b 2 a = 3 b 2 + b 6 = 0 a = 6 a = 1 nn v b + a = 3b + a = 3b = 3 b = 2Bi 2: a) ycbt tng ng vi PT x2 = (m +2)x m + 6 hay x2 - (m +2)x + m 6 = 0 c hai nghim dngphn bit. b) t t = 4 x + 2x 2Bi 3: a) x = 0, x = 1, x= -1 khng tha mn. Vi x khc cc gi tr ny, trc ht ta chng minh x phi l snguyn. +) x2 + x+ 6 l mt s chnh phng nn x2 + x phi l s nguyn. m +) Gi s x =vi m v n c c nguyn ln nht l 1. n m 2 m m 2 + mn Ta c x2 + x = 2 + = 2l s nguyn khi m 2 + mn chia ht cho n2 nn n thi vo lp 10 mn Ton nm 201263 64. www.VNMATH.comnn m + mn chia ht cho n, v mn chia ht cho n nn m2 chia ht cho n v do m v n c c nguyn2ln nht l 1, suy ra m chia ht cho n( mu thun vi m v n c c nguyn ln nht l 1). Do xphi l s nguyn.t x2 + x+ 6 = k2Ta c 4x2 + 4x+ 24 = 4 k2 hay (2x+1)2 + 23 = 4 k2 tng ng vi 4 k2 - (2x+1)2 = 23(x 3 + y3 ) (x 2 + y 2 ) x 2 (x 1) + y 2 (y 1)x2 y2 = = + (x 1)(y 1)(x 1)(y 1)y 1 x 1(x 1) + 2(x 1) + 1 (y 1) + 2(y 1) + 12 2= + y 1 x 1 (x 1) (y 1) 2(y 1) 2(x 1) 1221 =+ + x 1 + y 1 + y 1 + x 1 . y 1 x 1 Theo BT Csi(x 1) 2 (y 1) 2(x 1) 2 (y 1) 2 +2.= 2 (x 1)(y 1) y 1x 1 y 1x 12(y 1) 2(x 1) 2(y 1) 2(x 1)+ . =4 x 1 y 1 x 1 y 1 11 11+ 2 .y 1 x 1y 1 x 1 1 1 1 12 . + (x 1)(y 1) 2.2. . (x 1)(y 1) = 4 y 1 x 1y 1 x 1Bi 4 CE S P Q M N AO F Ba) Suy ra t hai tam gic ng dng l ABE v BSM AE MBb) T cu a) ta c = (1) AB BS thi vo lp 10 mn Ton nm 2012 64 65. www.VNMATH.comM MB = EM( do tam gic BEC vung ti E c M l trung im ca BC AE EMNn= AB BS C MOB = BAE, EBA + BAE = 900 , MBO + MOB = 900 Nn MBO = EBA do MEB = OBA( = MBE) Suy ra MEA = SBA (2)T (1) v (2) suy ra hai tam gic AEM v ABS ng dng(pcm.) c) D thy SM vung gc vi BC nn chng minh bi ton ta chng minh NP //SM. + Xt hai tam gic ANE v APB: T cu b) ta c hai tam gic AEM v ABS ng dng nn NAE = PAB , M AEN = ABP ( do t gic BCEF ni tip)AN AE Do hai tam gic ANE v APB ng dng nn=AP ABAM AE Li c = ( hai tam gic AEM v ABS ng dng)AS ABAM AN Suy ra =nn trong tam gic AMS c NP//SM( nh l Talet o)AS APDo bi ton c chng minh.Bi 5a. Gia s kt lun cua bai toan la sai, tc la trong ba i bt ky thi co hai i a u vi nhau ri. Gia si a gp cac i 2, 3, 4, 5. Xet cac b (1; 6; i) vi i {7; 8; 9;;12}, trong cac b nay phai co it nht mtcp a u vi nhau, tuy nhin 1 khng gp 6 hay i nn 6 gp i vi mi i {7; 8; 9;;12} , v ly vi i 6 nhth a u hn 4 trn. Vy co pcm. b. Kt lun khng ung. Chia 12 i thanh 2 nhom, mi nhom 6 i. Trong mi nhom nay, cho tt ca caci i mt a thi u vi nhau. Luc nay ro rang mi i a u 5 trn. Khi xet 3 i bt ky, phai co 2 i thuccung mt nhom, do o 2 i nay a u vi nhau. Ta co phan vi du.Co th giai quyt n gian hn cho cu a. nh sau:Do mi i a u 4 trn nn tn tai hai i A, B cha u vi nhau. Trong cac i con lai, vi A va B chiu 3 trn vi ho nn tng s trn cua A, B vi cac i nay nhiu nht la 6 va do o, tn tai i C trong s caci con lai cha u vi ca A va B. Ta co A, B, C la b ba i i mt cha u vi nhau. thi vo lp 10 mn Ton nm 201265 66. www.VNMATH.comTHI TUYN SINH VO LP 10 CHUYN TNH NG NAI NM HC 2012 - 2013 CHNH THC Mn thi: Ton chung Thi gian lm bi: 120 pht ( khng k thi gian giao ) ( thi ny gm mt trang, c bn cu)Cu 1: ( 2,5 im) . 1/ Gii cc phng trnh : a/ x 4 x 2 20 = 0 b/x +1 = x 1 x + y 3 =1 2/ Gii h phng trnh : y x =3 Cu 2 : ( 2,0 im) . Cho parabol y = x2 (P) v ng thng y = mx (d), vi m l tham s. 1/ Tm cc gi tr ca m (P) v (d) ct nhau ti im c tung bng 9. 2/ Tm cc gi tr ca m (P) v (d) ct nhau ti 2 im, m khong cch gia hai im ny bng 6Cu 3 : ( 2,0 im) 11 3 1 1/ Tnh : P = ( ).2 3 2+ 3 3 3 2/ Chng minh : a 5 + b5 a 3b 2 + a 2b3 , bit rng a + b 0 .Cu 4 : (3,5 im)Cho tam gic ABC vung A, ng cao AH. V ng trn tm O, ng knh AH, ng trn nyct cc cnh AB, AC theo th t ti D v E .1/ Chng minh t gic BDEC l t gic ni tip c ng trn.2/ Chng minh 3 im D, O, E thng hng.3/ Cho bit AB = 3 cm, BC = 5 cm. Tnh din tch t gic BDEC. --------HT------ thi vo lp 10 mn Ton nm 201266 67. www.VNMATH.com THI TUYN SINH VO LP 10 CHUYN TNH NG NAI NM HC 2012 - 2013 CHNH THC Mn thi: Ton ( mn chuyn) Thi gian lm bi: 150 pht ( khng k thi gian giao )( thi ny gm mt trang, c nm cu)Cu 1. (1,5 im) Cho phng trnh x 4 16 x 2 + 32 = 0 ( vi x R ) Chng minh rng x = 6 3 2 + 3 2 + 2 + 3 l mt nghim ca phng trnh cho.Cu 2. (2,5 im) 2 x( x + 1)( y + 1) + xy = 6 Gii h phng trnh ( vi x R, y R ). 2 y ( y + 1)( x + 1) + yx = 6Cu 3.(1,5 im) Cho tam gic u MNP c cnh bng 2 cm. Ly n im thuc cc cnh hoc pha trong tam gic u MNP sao cho khong cch gia hai im tu ln hn 1 cm ( vi n l s nguyn dng). Tm n ln nht tho mn iu kin cho.Cu 4. (1 im) Chng minh rng trong 10 s nguyn dng lin tip khng tn ti hai s c c chung ln hn 9.Cu 5. (3,5 im) Cho tam gic ABC khng l tam gic cn, bit tam gic ABC ngoi tip ng trn (I). Gi D,E,F ln lt l cc tip im ca BC, CA, AB vi ng trn (I). Gi M l giao im ca ng thng EF v ng thng BC, bit AD ct ng trn (I) ti im N (N khng trng vi D), gii K l giao im ca AI v EF. 1) Chng minh rng cc im I, D, N, K cng thuc mt ng trn. 2) Chng minh MN l tip tuyn ca ng trn (I). ----------HT----------- thi vo lp 10 mn Ton nm 2012 67 68. www.VNMATH.comGII THI VO LP 10CHUYN LNG TH VINH NG NAINM 2012 2013 Mn: Ton chung-----------------Cu 1: ( 2,5 im) . 1/ Gii cc phng trnh : a/ x 4 x 2 20 = 0 (*)t x 2 = t ;(t 0)(*) t2 t 20 = 0 (t1 = 5 (nhn) v t2 = - 4 ( loi)); Vi t = 5 => x2 = 5 x = 5Vy phng trnh c hai nghim x = 5 v x = - 5b/ x + 1 = x 1 ( iu kin x 1 )( x + 1) 2 = ( x 1) 2 x + 1 = x 2 2 x + 1 x 2 3 x = 0 x(x-3) = 0 x = 0 ( loi) v x = 3 ( nhn).Vy phng trnh c mt nghim x = 3. x + y 3 =1 2/ Gii h phng trnh : y x =3 T y x = 3 y 3 = x y 3 0 y 3 = y 31 x + y 3 =1 x + y 3 =1 x + y = 4 2 x =1 x = 2 (nhn) y x =3 y x =3 y x =3y = x + 3 y = 7 2 1 71 7Vy h phng trnh c 2 nghim (x; y): ( ; ), ( ; ) 2 22 2Cu 2 : ( 2,0 im) .x =01/ P.trnh honh giao im (P) v (d) : x 2 mx = 0 x( x m) = 0 1 x2 = mV giao im ( P) : y = x 2 y = m 2 . Vi y = 9 => m2 = 9 (m = 3 v m = -3)Vy vi m = 3 th (P) v (d) ct nhau ti im c tung bng 9.2/ T cu 1 => (P) v (d) lun ct nhau ti hai im phn bit khi m 0 . thi vo lp 10 mn Ton nm 201268 69. www.VNMATH.comKhi giao im th nht l gc to O ( x = 0; y = 0), giao im th 2 l im A c ( x = m; y = m2).Khong cch gia hai giao im : AO = m 2 + m 4 = 6 m 4 + m 2 6 = 0 (1)t t = m 2 ;(t 0) (1) t 2 + t 6 = 0 (t1 = 3 ( nhn ) v t2 = - 2 ( loi))Vi t1 = 3 m2 = 3 , m = 3 ( nhn)Vy vi m = 3 th (P) ct (d) ti hai im c khong cch bng 6 .Cu 3 : ( 2,0 im)1/ Tnh:1 1 3 1 2 + 3 2 + 3 3 1P=( ). =.=2 2 3 2+ 3 3 3 433( 3 1)2/ Ta c:a 5 + b5 a 3b 2 + a 2b3 a 5 + b5 a 3b 2 a 2b3 0 a 3 (a 2 b 2 ) b 3 (a 2 b 2 ) 0 (a 3 b3 )(a 2 b 2 ) 0 (a b) 2 (a + b)(a 2 + b 2 + ab) 0V :( a b) 2 0 (vi mi a, b R ).a+b 0 ( theo gi thit) a 2 + b 2 + ab 0 ( vi mi a, b R )Nn bt ng thc cui ng. Vy a 5 + b5 a 3b 2 + a 2b3 vi a + b 0 (pcm)Cu 4 : (3,5 im)A EO D BCH 1/ Ni H vi E . + HEA = 900 ( v AH l ng knh), AHC = 900 ( AH l ng cao)=> AHE = ACB (cng ph vi EHC )(1) + ADE = AHE ( gc ni tip cng chn cung AE)(2) T (1) v (2) => ADE = ACB =>T gic BDEC ni tip ng trn ( c gc i bng gc k b gc i) 2/ V DAE = 900 => DE l ng knh => D, O, E thng hng (pcm). 3/ Ta c S BDEC = S ABC S ADE + ABC vung c AH l ng cao:AB. ACAC = BC 2 AB 2 = 4cm => sABC = = 6 (cm2) 2 AB. AC 12 DE = AH = = (cm) ( cng l ng knh t O). BC5 + ADE v ABC c : A chung , ADE = ACB ( cu 1) => ADE ~ ABC (g.g) => t s din tch bng bnh phng t ng dng : 2 S AED DE S.DE 2 = S AED = ABC 2 S ABC BC BC thi vo lp 10 mn Ton nm 2012 69 70. www.VNMATH.com DE 2122+ S BDEC = S ABC S ADE = S ABC (1 ) = 6(1 2 2 ) = 4,6176 (cm2) BC 2 5 .5---------HT---------GII THI VO LP 10CHUYN LNG TH VINH NG NAINM 2012 2013Mn: Ton chuyn-----------------Cu 1: Phng trnh cho : x 16 x + 32 = 0 ( vi x R ) ( x 2 8) 2 32 = 0 (1)42Vi x = 6 3 2 + 3 2 + 2 + 3 x = 3 2 2 + 3 2 + 2 + 3 => x 2 = 8 2 2 + 3 2 3 2 3Th x vo v phi ca (1) ta c:( x 2 8) 2 32 = (8 2 2 + 3 2 3 2 3 8) 2 32 = 4(2 + 3) + 4 3 + 12(2 3) 32= 8 + 4 3 + 8 3 + 24 12 3 32 = 0 ( v phi bng v tri)Vy x = 6 3 2 + 3 2 + 2 + 3 l mt nghim ca phng trnh cho ( pcm) 2 x( x + 1)( y + 1) + xy = 6 (1) 2 x( x + 1)( y + 1) = 6 xyCu 2: H pt cho 2 y ( y + 1)( x + 1) + yx = 6 (2) 2 y ( y + 1)( x + 1) = 6 xyThay x = 0, y = 0 th h khng tho . Thay x = -1 v y = -1 vo, h khng tho =>( x; y ) (0;0); xy 0; x + 1 0; y + 1 0 6 xy 0(*)x 6 xy- Chia tng v ca hai phng trnh cho nhau : => = xy ( x y ) = 6( x + y )y 6 xyThay x = y, h pt c v phi bng nhau, v tri khc nhau (khng tho) => x y 0 ) (**) 6( x + y )=> xy =(3) x y- Cng tng v (1) v (2) ca h ta c pt: 2(x+y)(x+1)(y+1) + 2xy = 0(4)6( x + y ) 6( x + y ) (x + y) ( x + y + xy + 1) + xy = 0 ( x + y )( x + y + 1 + )+ =0x y x y thi vo lp 10 mn Ton nm 2012 70 71. www.VNMATH.com x+ y = 0 6( x + y + 1)6 x + y +1 = 0 ( x + y )( x + y + 1 + ) = 0 ( x + y )( x + y + 1)(1 +) = 0 x yx y6 1 + x y = 0 - Vi x + y = 0 x = - y. Th vo h => -2y2 = 0 (y = 0 v x = 0) khng tho (*)- Vi x + y +1 =0 x = -y - 1 th vo phng trnh (1) ca h ta c : y + 2 = 0 y = 2 2 y 3 + 3 y 2 + y + 6 = 0 ( y + 2)(2 y 2 y + 3) = 0 2 2 y y + 3 = 0(vn) Vi y = - 2 => x = 1.Th vo h tho, vy c nghim 1: (x; y) = (1; - 2) 6- Vi 1 + = 0 x y+6 = 0 x = y6 x yTh x = y -6 vo pt (2) ca h : 2 y +1 = 0 (2) 2 y 7 y 16 y 6 = 0 (2 y + 1)( y 4 y 6) = 0 232 2 y 4y 6 = 0 y1 = 2 + 10y2 - 4y - 6 = 0 y2 = 2 101 2y +1 = 0 y3 = 2 x1 = 4 + 10 T ba gi tr ca y trn ta tm c ba gi tr x tng ng: x2 = 4 10 13 x3 = 2Th cc gi tr (x; y) tm c vo h (tho). Vy h phng trnh cho c 4 nghim ( x;y): 13 1(1; -2), ( 4 + 10; 2 + 10), (4 10; 2 10), ( ; ).2 2Cu 3. (Cch 1)Tam gic u c cnh bng 2 cm th din tch bng 3 cm2 , tam gic u c cnh bng 1 cm th din 33tch bng cm2 . Nu tam gic u c cnh > 1cm th din tch >cm2 4 4Gi t l s tam gic u c cnh bng > 1cm cha c trong tam gic u c cnh 2 cm: 1 t p 4 ( vi t l s nguyn dng) => tmax = 3.Theo nguyn l Drichen s c 1 trong t tam gic u c cnh > 1cm cha ti a 2 im tho mn khongcch gia hai im bt k lun > 1 cm.Vy s im tho yu cu bi ton l : 2 n 4 Vy nmax = 4(Cch 2): Gii theo kin thc hnh hc thi vo lp 10 mn Ton nm 2012 71 72. www.VNMATH.com Nu ta chn 3 im 3 nh ca tam gic u cnh bng 2 cm v 3 ng trn ng knh 1 cm, ccng trn ny tip xc vi nhau trung im mi cnh tam gic. => Cc im khc trong tam gic cch 3nh > 1cm ch c th nm trong phn din tch cn li ca tam gic (ngoi phn din tch b ba hinh trn cheph), c gii hn bi 3 cung trn bn kinh 1 cm. V 3 dy cung l 3 ng trung bnh ca tam gic c di 1 cm => khong cch gia hai im bt knm trong phn din tch cn li ca tam gic lun 1 cm. => trong phn din tch ch ly c 1 im m khong cch n 3 nh ca tam gic lun > 1 cm. Vy s im ln nht tho mn khong cch gia hai im bt k > 1cm l :nmax = 3 + 1 = 4 im.Cu 4. Gi a v b l hai s bt k trong 10 s nguyn dng lin tip vi a > b ( a; b nguyn dng)1 a b 9 .Gi n l c chung ca a v b, khi : a = n.x v b = n.y ( n, x, y l s nguyn dng).1 99V a > b => x > y => x y 1 1 n.x n. y 9 x y 1 n 9n nnVy trong 10 s nguyn dng lin tip khng tn ti hai s c c chung ln hn 9.Cu 5. A E N K F I M B DC1)Ni N v F, D v F.- Xt ANF v AFD c: AFN = ADF ( v AF l tt) v FAD chung => ANF AFD (g.g) => AN AF = AF2 = AN . AD(1) AF AD- Xt AFI c: AF IF ( v AF tip tuyn, FI l bn knh) v FK AI ( v AF v AE tt chung v AI ni tm)=> AFI vung ti F c FK l ng cao) => AK.AI = AF2 (2)- Xt ANK v AID c: thi vo lp 10 mn Ton nm 2012 72 73. www.VNMATH.com + IAD chung.AN AI + T (1) v (2) => AN.AD = AK.AI => =AK AD=> ANK AID (c.g.c) => NKA = IDN (3)- T (3) => t gic DIKN ni tip t (v c gc i bng gc k b gc i)=> cc im I,D,N,K cng thuc mt ng trn. (pcm).2) Ta c ID DM ( DM l tip tuyn, DI l bn knh) v IK KM ( cu 1) => t gic DIKM ni tip ngtrn ng knh MI. V 4 im D, I, K, N cng thuc mt ng trn ( cu 1) => hai ng trn ny cngngoi tip DIK => hai ng trn trng nhau => N cng nm trn ng trn ng knh MI => INM =900 . V IN l bn knh ng trn (I), MN IN => MN l tip tuyn ca ng trn (I) ti tip im N. (pcm). -----------HT---------- www.VNMATH.com thi vo lp 10 mn Ton nm 2012 73 74. www.VNMATH.com CHNH THC thi vo lp 10 mn Ton nm 201274 75. www.VNMATH.comGI GII:Cu 1c C = 1Cu 2a ( 2;1) ; Cu 2b b = - 1Cu 3a a = 1Cu 3b A ( -1 ; 1 ) ; B (2 ; 4 )Cu 4a1 = 12 > 0 ; nn pt lun co 2 nghim phn bit vi moi x Cu 4 a2 => x1 + x2 = - 5 ; x1x2 = 3Cu 4bGoi x ( km/h) la vt xe II => vt xe I la x + 10 ( km/h ) ; x> 0100Th gian xe I i ht qg : (h) x100Th gian xe II i ht qg :(h) x + 10100 1001PT - = => x = 40xx + 102KL Cu 5 : a1. MH = 20 ( cm ) ; ME = 12 ( cm)2. NPFE la h thang cnb)b1b2 Tam giac ABC vung tai A co AH la g cao => AB2 = BH.BC (1) BH BETam giac BHE g dang vi tam giac BDC =>= => BH .BC = BD.BE (2) BD BCT (1) va (2) => AB2 = BD . BE thi vo lp 10 mn Ton nm 2012 75 76. www.VNMATH.com S GIO DC - O TOK THI TUYN SINH VO LP 10 THPT CHUYNTNH NINH BNHMn thi: TON CHNH THCNgy thi: 26 / 6 / 2012Thi gian lm bi: 120 phtCu 1 (2 im). Cho phng trnh bc hai n x, tham s m: x2 + 2mx 2m 3 = 0 (1) a) Gii phng trnh (1) vi m = -1. b) Xc nh gi tr ca m phng trnh (1) c hai nghim x1, x2 sao cho x12 + x 2 nh nht. Tm2nghim ca phng trnh (1) ng vi m va tm c.Cu 2 (2,5 im). 6x + 4 3x 1 + 3 3 x 3 1. Cho biu thc A= 3x 3 x + 2 3 x + 4 1 + 3 x 3 3x 83 a) Rt gn biu thc A.b) Tm cc gi tr nguyn ca x biu thc A nhn gi tr nguyn.2. Gii phng trnh: x + 1 x + x(1 x ) = 1Cu 3 (1,5 im). Mt ngi i xe p t A ti B, qung ng AB di 24 km. Khi i t B tr v A ngi tng vn tc thm 4 km/h so vi lc i, v vy thi gian v t hn thi gian i l 30 pht. Tnh vn tc ca xep khi i t A ti B.Cu 4 (3 im). Cho ABC nhn ni tip (O). Gi s M l im thuc on thng AB (M A, B); N l imthuc tia i ca tia CA sao cho khi MN ct BC ti I th I l trung im ca MN. ng trn ngoi tip AMN ct (O) ti im P khc A.1. C MR cc t gic BMIP v CNPI ni tip c.2. Gi s PB = PC. Chng minh rng ABC cn. xCu 5 (1 im). Cho x; y R , tha mn x2 + y2 = 1. Tm GTLN ca : P = y + 2HNG DN GII:www.VNMATH.com2) Gii pt : x + 1x +x (1 x ) =1 K : 0 x 1t x = a 0; 1 x = b 0 a + b + ab = 1(*)Ta c a 2 + b 2 = 1(**)T tm c nghim ca pt l x = 0Cu 5 :T x 2 + y 2 = 1 1 x, y 1 2 1 y +2 1 + 2 xV P = y + 2 x = P( y + 2 ) thay vo x 2 + y 2 = 1a v pt: ( P 2 +1) y 2 + 2 2 P 2 y + 2 P 2 1 = 0 2 x = 2Dng iu kin c nghim ca pt bc hai P 1 PMax =1 y = 2 2 thi vo lp 10 mn Ton nm 2012 76 77. www.VNMATH.comS GIO DC V O TO K THI TUYN SINH VO 10 - THPTTNH LO CAI NM HC: 2012 2013MN: TON CHNH THC Thi gian: 120 pht (khng k thi gian giao )Cu I: (2,5 im)() ( ) 2 31. Thc hin php tnh: a) 3 2 10 36 + 64b) 2 3 + 3 2 5 .2a 2 + 4 1 12. Cho biu thc: P = 1 a3 1 + a 1 aa) Tm iu kin ca a P xc nhb) Rt gn biu thc P.Cu II: (1,5 im)1. Cho hai hm s bc nht y = -x + 2 v y = (m+3)x + 4. Tm cc gi tr ca m th ca hm s cho l: a) Hai ng thng ct nhau b) Hai ng thng song song.2. Tm cc gi tr ca a th hm s y = ax2 (a 0) i qua im M(-1; 2).Cu III: (1,5 im)1. Gii phng trnh x 2 7x 8 = 02. Cho phng trnh x2 2x + m 3 = 0 vi m l tham s. Tm cc gi tr ca m phng trnh c hai nghimx1; x2 tha mn iu kin x1 x 2 + x1x 2 = 6 33Cu IV: (1,5 im)3x 2y = 11. Gii h phng trnh . x + 3y = 22x y = m 12. Tm m h phng trnh c nghim (x; y) tha mn iu kin x + y > 1.3x + y = 4m + 1Cu V: (3,0 im) Cho na ng trn tm O ng knh AB = 2R v tip tuyn Ax cng pha vi na ngtrn i vi AB. T im M trn Ax k tip tuyn th hai MC vi na ng trn (C l tip im). AC ct OMti E; MB ct na ng trn (O) ti D (D khc B).a) Chng minh AMOC l t gic ni tip ng trn.b) Chng minh AMDE l t gic ni tip ng trn.c) Chng mnh ADE = ACO -------- Ht ---------HNG DN GII:Cu I: (2,5 im)1. Thc hin php tnh:a) 3 2 10 36 + 64 = 3 8 100 = 2 10 = 12 () ()23b)2 3 +3 2 5 =2 3 + 2 5 = 3 2 + 2 5 = 22a 2 + 4 1 12. Cho biu thc: P = 1 a3 1 + a 1 aa) Tm iu kin ca a P xc nh:P xc nh khi a 0 v a 1b) Rt gn biu thc P. thi vo lp 10 mn Ton nm 201277 78. www.VNMATH.comP= 2a 2 + 411 = () ( 2a + 4 1 a ( a 2 + a + 1) 1 + a ( a 2 + a + 1) 2)1 a3 1 + a 1 a ( 1 a ) ( a 2 + a + 1)2a 2 + 4 a 2 a 1 + a 2 a + a a + a a 1 a 2 a a a a= ( 1 a ) ( a 2 + a + 1)2 2a2=( 1 a ) ( a + a + 1) a + a + 1 2= 22Vy vi a 0 v a 1 th P = a + a +1 2Cu II: (1,5 im)1. Cho hai hm s bc nht y = -x + 2 v y = (m+3)x + 4. Tm cc gi tr ca m th ca hm s cho l: a) hm s y = (m+3)x + 4 l hm s bc nht th m + 3 0 suy ra m -3. th ca hai hm s cho l hai ng thng ct nhau a a -1 m+3 m -4Vy vi m -3 v m -4 th th ca hai hm s cho l hai ng thng ct nhau. b) th ca hm s cho l Hai ng thng song song a = a 1 = m + 3 m = 4 tha mn iu kin m -3 b b 2 4Vy vi m = -4 th th ca hai hm s cho l hai ng thng song song.2. Tm cc gi tr ca a th hm s y = ax2 (a 0) i qua im M(-1; 2).V th hm s y = ax2 (a 0) i qua im M(-1; 2) nn ta thay x = -1 v y = 2 vo hm s ta c phng trnh2 = a.(-1)2 suy ra a = 2 (tha mn iu kin a 0)Vy vi a = 2 th th hm s y = ax2 (a 0) i qua im M(-1; 2).Cu III: (1,5 im)1. Gii phng trnh x 2 7x 8 = 0 c a b + c = 1 + 7 8 = 0 suy ra x1= -1 v x2= 82. Cho phng trnh x2 2x + m 3 = 0 vi m l tham s. Tm cc gi tr ca m phng trnh c hai nghimx1; x2 tha mn iu kin x1 x 2 + x1x 2 = 6 . 3 3 phng trnh c hai nghim x1; x2 th 0 1 m + 3 0 m 4Theo viet ta c: x1+ x2 =2 (1) v x1. x2 = m 3 (2)x1 x 2 + x1x 3 = 6 x1x 2 ( x1 + x 2 ) 2x1x 2 = 6 (3) 32Theo u bi:2Th (1) v (2) vo (3) ta c: (m - 3)(2)2 2(m-3)=6 2m =12 m = 6 Khng tha mn iu kin m 4 vykhng c gi tr no ca m phng trnh c hai nghim x1; x2 tha mn iu kin x1 x 2 + x1x 2 = 6 .3 3Cu IV: (1,5 im) 3x 2y = 13 ( 3y 2 ) 2y = 1 7y = 7y = 11. Gii h phng trnh . x + 3y = 2 x = 3y 2 x = 3y 2 x = 12x y = m 12. Tm m h phng trnh c nghim (x; y) tha mn iu kin x + y > 1.3x + y = 4m + 12x y = m 15x = 5mx = m x = m 3x + y = 4m + 1 2x y = m 1 2m y = m 1 y = m + 1 thi vo lp 10 mn Ton nm 201278 79. www.VNMATH.comM x + y > 1 suy ra m + m + 1 > 1 2m > 0 m > 0.Vy vi m > 0 th h phng trnh c nghim (x; y) tha mn iu kin x + y > 1.Cu V: (3,0 im) Cho na ng trn tm O ng knh AB = 2R v tip tuyn Ax cng pha vi na ngtrn i vi AB. T im M trn Ax k tip tuyn th hai MC vi na ng trn (C l tip im). AC ct OMti E; MB ct na ng trn (O) ti D (D khc B).a) Chng minh AMCO l t gic ni tip ng trn.b) Chng minh AMDE l t gic ni tip ng trn.c) Chng mnh ADE = ACO MGii.a) MAO = MCO = 90 0 nn t gic AMCO ni tip DC b) MEA = MDA = 900 . T gic AMDE cD, E cng nhn AM di cng mt gc 900ENn AMDE ni tip c) V AMDE ni tip nn ADE = AME cng chan cung AE AB O V AMCO ni tip nn ACO = AME cng chan cung AO Suy ra ADE = ACO thi vo lp 10 mn Ton nm 201279 80. www.VNMATH.com S GIO DC V O TO K THI TUYN SINH VO LP 10 CHUYNGIA LAI Nm hc 2012 2013 chnh THC CHNH thcMn thi: Ton (khng chuyn)Ngy thi: 26/6/2012 Thi gian lm bi: 120 phtCu 1. (2,0 im) x 2 Cho biu thc Q = x +2 x + 2 x +1 x 1 () x + x , vi x > 0, x 1 a. Rt gn biu thc Q b. Tm cc gi tr nguyn ca x Q nhn gi tr nguyn.Cu 2. (1,5 im) Cho phng trnh x 2 2(m + 1)x + m 2 = 0 , vi x l n s, m R a. Gii phng trnh cho khi m = 2 b. Gi s phng trnh cho c hai nghim phn bit x1 v x 2 . Tm h thc lin h gia x1 v x 2 mkhng ph thuc vo m.Cu 3. (2,0 im)(m + 1)x (m + 1)y = 4m Cho h phng trnh , vi m Rx + (m 2)y = 2 a. Gii h cho khi m = 3 b. Tm iu kin ca m phng trnh c nghim duy nht. Tm nghim duy nht .Cu 4. (2,0 im) Cho hm s y = x 2 c th (P). Gi d l ng thng i qua im M(0;1) v c h s gc k. a. Vit phng trnh ca ng thng d b. Tm iu kin ca k t d ct th (P) ti hai im phn bit.Cu 5. (2,5 im) Cho tam gic nhn ABC (AB < AC < BC) ni tip trong ng trn (O). Gi H l giao im ca haing cao BD v CE ca tam gic ABC (D AC, E AB) a. Chng minh t gic BCDE ni tip trong mt ng trn b. Gi I l im i xng vi A qua O v J l trung im ca BC. Chng minh rng ba im H, J, Ithng hng 1 11 c. Gi K, M ln lt l giao im ca AI vi ED v BD. Chng minh rng2 = 2 +DK DADM 2 HNG DN GII: Cu 1. x 2 a. Q = x +2 x + 2 x +1 x 1 (x+ x)= x +2 x 2 x () x +1 ( ) ( )()2 x +1 x 1x +1 x +2 x 2 x + 1+ 1 x 1 1 1 1 = x = x = 1+ 1+ x x +1 x 1 x +1x 1 x +1x 1 thi vo lp 10 mn Ton nm 201280 81. www.VNMATH.com 1 1 x 1+ x + 1 2 x 2x = + x =. x =. x =x +1x 1 x 1 x 1 x 12x Vy Q= x 1 b.Q nhn gi tr nguyn2x 2x 2 + 2 2Q= == 2+ x 1x 1 x 1 2Q khi khi 2 chia ht cho x 1 x 1x = 0x = 2 x 1 = 1x = 2 i chiu iu kin thx = 3 x 1 = 2 x = 1x = 3 Cu 2. Cho pt x 2 2(m + 1)x + m 2 = 0 , vi x l n s, m R a. Gii phng trnh cho khi m = 2Ta c phng trnh x 2 + 2x 4 = 0 ( 5)2x 2 + 2x 4 = 0 x 2 + 2x + 1 = 5 ( x + 1) = 5 =2 x + 1 = 5 x = 1 5 x +1 = 5 x + 1 = 5 x = 1 + 5 Vy phng trinh c hai nghim x = 1 5 v x = 1 + 5 b. x1 + x 2 = 2m + 2 (1) x + x 2 = 2m + 2Theo Vi-et, ta c 1 x1x 2 = m 2 (2) m = x1 x 2 + 2 x + x 2 = 2 ( x1 x 2 + 2 ) + 2 1 m = x1 x 2 + 2Suy ra x1 + x 2 = 2 ( x1x 2 + 2 ) + 2 x1 + x 2 2x1x 2 6 = 0 (m + 1)x (m + 1)y = 4m Cu 3. Cho h phng trnh , vi m R x + (m 2)y = 2 a. Gii h cho khi m = 3 2x + 2y = 12 x + y = 6 x = 7 Ta c h phng trnh x 5y = 2 x 5y = 2y = 1 Vy h phng trnh c nghim ( x; y ) vi ( 7;1) b. iu kin c nghim ca phng trnh m + 1 ( m + 1) ( m + 1) ( m 2 ) ( m + 1)1 m2 ( m + 1) ( m 2 ) + ( m + 1) 0 ( m + 1) ( m 1) 0m + 1 0m 1 m 1 0m 1 Vy phng trnh c nghim khi m 1 v m 1 thi vo lp 10 mn Ton nm 201281 82. www.VNMATH.com(m + 1)x (m + 1)y = 4m m 1 Gii h phng trnh khi x + (m 2)y = 2 m 1 4m 4m 2 (m + 1)x (m + 1)y = 4mx y =4m x = y + m + 1 x = m +1 m +1 . x + (m 2)y = 2x + (m 2)y = 2 y = 2y =2 m +1 m +1 4m 2 2 Vy h c nghim (x; y) vi ; m + 1 m + 1 Cu 4. a. Vit phng trnh ca ng thng d ng thng d vi h s gc k c dng y = kx + b ng thng d i qua im M(0; 1) nn 1 = k.0 + b b = 1 Vy d : y = kx + 1 b. Phng trnh honh giao im ca (P) v d x 2 = kx + 1 x 2 + kx + 1 = 0 , c = k 2 4 d ct (P) ti hai im phn bit khi > 0 k < 2 k 2 4 > 0 k 2 > 4 k 2 > 22 k > 2 k > 2 Cu 5. a.BCDE ni tip BEC = BDC = 900 Suy ra BCDE ni tip ng trn ng knh BC b.H, J, I thng hng IB AB; CE AB (CH AB) Suy ra IB // CH IC AC; BD AC (BH AC) Suy ra BH // IC Nh vy t gic BHCI l hnh bnh hnh J trung im BC J trung im IH Vy H, J, I thng hng 1 c.ACB = AIB = AB 2 ACB = DEA cng b vi gc DEB ca t gic ni tip BCDE BAI + AIB = 900 v ABI vung ti B Suy ra BAI + AED = 900 , hay EAK + AEK = 900 Suy ra AEK vung ti K Xt ADM vung ti M (suy t gi thit) DK AM (suy t chng minh trn)www.VNMATH.1 1 1 Nh vy2 =2 +DKDA DM 2 thi vo lp 10 mn Ton nm 201282 83. www.VNMATH.comS GIO DC V O TO K THI TUYN SINH LP 10 THPTTHANH HANM HC 2012-2013Mn thi : Ton THI CHNH THC Thi gian : 120 pht khng k thi gian giao A Ngy thi 29 thng 6 nm 2012Bi 1: (2.0 im) 1- Gii cc phng trnh sau : a) x - 1 = 0 b) x2 - 3x + 2 = 02 x y = 72- Gii h phng trnh : x+y =2 1 1 a 2 +1Bi 2: (2.0 im) Cho biu thc : A =2+2 a+ 2 2 a - 1a2 1- Tm iu kin xc nh v rt gn biu thc A 1 2- Tm gi tr ca a ; bit A < 3Bi 3: (2.0 im)1- Cho ng thng (d) : y = ax + b .Tm a; b ng thng (d) i qua im A( -1 ; 3) vsong song vi ng thng (d) : y = 5x + 32- Cho phng trnh ax2 + 3(a + 1)x + 2a + 4 = 0 ( x l n s ) .Tm a phmg trnh cho c hai nghim phn bit x1 ; x2 tho mn x12 + x 2 = 42Bi 4: (3.0 im) Cho tam tam gic u ABC c ng cao AH . Trn cnh BC ly im Mbt k ( M khng trng B ; C; H ) T M k MP ; MQ ln lt vung gc vi cc cnh AB ; AC( P thuc AB ; Q thuc AC) 1- Chng minh :T gic APMQ ni tip ng trn 2- Gi O l tm ng trn ngoi tip t gic APMQ .Chng minh OH PQ3- Chng minh rng : MP +MQ = AHBi 5: (1.0 im) Cho hai s thc a; b thay i , tho mn iu kin a + b 1 v a > 08a 2 + bTm gi tr nh nht ca biu thc A=+ b24a ---------------------------------------HT ---------------------------------- thi vo lp 10 mn Ton nm 201283 84. www.VNMATH.com thi vo lp 10 mn Ton nm 201284 85. www.VNMATH.com thi vo lp 10 mn Ton nm 201285 86. www.VNMATH.com thi vo lp 10 mn Ton nm 201286 87. www.VNMATH.com thi vo lp 10 mn Ton nm 201287 88. www.VNMATH.com thi vo lp 10 mn Ton nm 201288 89. www.VNMATH.com S GIO DC V O TO K THI TUYN SINH LP 10 THPT QUNG NINHNM HC 2012 2013 CHNH THCMN: TON(Dng cho mi th sinh d thi)Ngy thi: 28/6/2012 Thi gian lm bi: 120 pht (Khng k thi gian giao ) ( thi ny c 01 trang) Cu I. (2,0 im)1) Rt gn cc biu thc sau: 111 2a) A = 2 + 18 b) B =+ vi x 0, x 1 2 x 1 x +1 x 1 2x + y = 5 2. Gii h phng trnh: x + 2 y = 4Cu II. (2,0 im) Cho phng trnh (n x): x2 ax 2 = 0 (*) 1. Gii phng trnh (*) vi a = 1. 2. Chng minh rng phng trnh (*) c hai nghim phn bit vi mi gi tr ca a. 3. Gi x1, x2 l hai nghim ca phng trnh (*). Tm gi tr ca a biu thc: N= x1 + ( x1 + 2)( x2 + 2) + x2 c gi tr nh nht. 2 2Cu III. (2,0 im)Gii bi ton bng cch lp phng trnh hoc h phng trnh.Qung ng sng AB di 78 km. Mt chic thuyn my i t A v pha B. Sau 1 gi, mt chic can i t B v pha A. Thuyn v ca n gp nhau ti C cch B 36 km. Tnh thi gian ca thuyn, thi gian ca can i t lc khi hnh n khi gp nhau, bit vntc ca ca n ln hn vn tc ca thuyn l 4 km/h. Cu IV. (3,5 im)Cho tam gic ABC vung ti A, trn cnh AC ly im D (D A, D C). ng trn (O) ng knh DC ct BC ti E (E C). 1. Chng minh t gic ABED ni tip. 2. ng thng BD ct ng trn (O) ti im th hai I. Chng minh ED l tia phn gic ca gc AEI. 3. Gi s tg ABC = 2 Tm v tr ca D trn AC EA l tip tuyn ca ng trn ng knh DC.CuV. (0.5 im) Gii phng trnh:7 + 2 x x = (2 + x ) 7 xHNG DN GII:Cu IV :c. EA l tip tuyn ca .Trn, . knh CD th gc E 1 = gc C1 (1)M t gic ABED ni tip nn gc E1 = gc B1 (2)T (1) v (2) gc C1 = gc B1 ta li c gc BAD chung nn AB ADAB 2 ABD ACB = AB2 = AC.AD AD = (I) AC ABACAC AB 1Theo bi ra ta c : tan (ABC) = = 2 nn( II ) ABAC 2 thi vo lp 10 mn Ton nm 201289 90. www.VNMATH.comABT (I) v (II) AD = . 2ABVy AD =th EA l tip tuyn ca T, knh CD2Cu V:Gii phng trnh: 7 + 2 x x = (2 + x ) 7 xt 7 x = t ; x = v K v, t 0 t 2 + 2v = (2 + v).t ... (t v)(t 2) = 0 t = v hoc t=2Nu t= 2 th 7 x =2 x = 3 (TM)Nu t = v th7 x =x x = 3,5 thi vo lp 10 mn Ton nm 201290 91. www.VNMATH.comS GIO DC V O TO K THI TUYN SINH VO LP 10 THPTKHNH HANM HC 2011 - 2012Mn thi: TON CHNH THC Ngy thi : 21/06/2011 Thi gian lm bi: 120 phtBi 1( 2 im) 2+ 3+ 6+ 8+4 1)n gin biu thc: A =2+ 3+ 411 P = a();(a 1) 2) Cho biu thc: a a 1 a + a 1Rt gn P v chng t P 0Bi 2( 2 im) 1) Cho phng trnh bc hai x2 + 5x + 3 = 0 c hai nghim x1; x2. Hy lp mt phng trnh bc hai c hainghim (x12 + 1 ) v ( x22 + 1). 23 x + y2 = 4 2) Gii h phng trnh 41 =1 x y2 Bi 3( 2 im)Qung ng t A n B di 50km.Mt ngi d nh i xe p t A n B vi vn tc khng i.Khi i c2 gi,ngi y dng li 30 pht ngh.Mun n B ng thi gian nh,ngi phi tng vn tc thm 2km/h trn qung ng cn li.Tnh vn tc ban u ca ngi i xe p.Bi 4( 4 im) Cho tam gic ABC c ba gc nhn v H l trc tm.V hnh bnh hnh BHCD.ng thng i qua D v songsong BC ct ng thng AH ti E.1) Chng minh A,B,C,D,E cng thuc mt ng trn2) Chng minh BAE = DAC3) Gi O l tm ng trn ngoi tip tam gic ABC v M l trung im ca BC,ng thng AM ct OHti G.Chng minh G l trng tm ca tam gicABC.4) Gi s OD = a.Hy tnh di ng trn ngoi tip tam gic BHC theo a HNG DN GII:Bi 1 2 + 3 + 2 + 6 + 8 + 2 ( 2 + 3 + 4)(1 + 2) 3) A == = 1+ 2 2+ 3+ 4 2+ 3+ 4a + a 1 a + a 1P = a( ); a 1a a +1 4)= a 2 a 1 = a 1 2 a 1 + 1; vi : a 1 P = ( a 1 1) 2 0; a 1 thi vo lp 10 mn Ton nm 2012 91 92. www.VNMATH.com2Bi 2 x + 5x + 3 = 0 1) C = 25 12 = 13 > 0 Nn pt lun c 2 nghim phn bit x1+ x2 = - 5 ; x1x2 = 3 Do S = x12 + 1 + x22 + 1 = (x1+ x2)2 - 2 x1x2 + 2 = 25 6 + 2 = 21V P = (x12 + 1) (x22 + 1) = (x1x2)2 + (x1+ x2)2 - 2 x1x2 + 1 = 9 + 20 = 29 Vy phng trnh cn lp l x2 21x + 29 = 0 2) K x 0; y 22 3 14x y 2 = 4 + =7x = 2x = 2x 3 12 3 = 3 2 + 3 = 41 + y 2 = 4 y = 3 x y2x y2 Vy HPT c nghim duy nht ( x ;y) = ( 2 ;3)Bi 3 : Gi x(km/h) l vtc d nh; x > 0 ; c 30 pht = (h) 50 Th gian d nh :( h)x Qung ng i c sau 2h : 2x (km) Qung ng cn li : 50 2x (km) Vn tc i trn qung ng cn li : x + 2 ( km/h)50 2 x Th gian i qung ng cn li :( h)x+2 1 50 2 x 50 2+ + = Theo bi ta c PT: 2x+2x Gii ra ta c : x = 10 (tha K bi ton)Vy Vn tc d nh : 10 km/hBi 4 :A Gii cu c) V BHCD l HBH nn H,M,D thng hng Tam gic AHD c OM l TBnh => AH = 2 OM V AH // OM H 2 tam gic AHG v MOG c HAG = OMG ( slt )GOAGH = MGO ( ) AHGMOG (G G )BCMAH AG= =2MO MG Hay AG = 2MGED Tam gic ABC c AM l trung tuyn; G AM Do G l trng tm ca tam gic ABC d) BHC = BDC ( v BHCD l HBH) c B ;D ;C ni tip (O) bn knh l a Nn tam gic BHC cng ni tip (K) c bn knh a Do C (K) = 2 a ( VD) thi vo lp 10 mn Ton nm 2012 92 93. www.VNMATH.comS GIO DC-O TOK THI TUYN SINH VO 10 THPT NM 2012 BNH NH Kha ngy 29 thng 6 nm 2012 CHNH THCMn thi: TON Ngy thi: 30/6/2012Thi gian lm bi: 120 pht (khng k thi gian giao )Bi 1: (3, 0 im)Hc sinh khng s dng my tnh b tia) Gii phng trnh: 2x 5 = 0 y x = 2b) Gii h phng trnh: 5x 3y = 105 a 33 a +1a2 + 2 a + 8 c) Rt gn biu thc A = + vi a 0, a 4 a 2a +2 a4d) Tnh gi tr ca biu thc B = 4 + 2 3 + 7 4 3Bi 2: (2, 0 im) Cho parabol (P) v ng thng (d) c phng trnh ln lt l y = mx 2 vy = ( m 2 ) x + m 1 (m l tham s, m 0). a) Vi m = 1 , tm ta giao im ca (d) v (P). b) Chng minh rng vi mi m 0 ng thng (d) lun ct parabol (P) ti hai im phn bit.Bi 3: (2, 0 im)Qung ng t Quy Nhn n Bng Sn di 100 km. Cng mt lc, mt xe my khi hnh t QuyNhn i Bng Sn v mt xe t khi hnh t Bng Sn i Quy Nhn. Sau khi hai xe gp nhau, xe my i 1gi 30 pht na mi n Bng Sn. Bit vn tc hai xe khng thay i trn sut qung ng i v vn tc caxe my km vn tc xe t l 20 km/h. Tnh vn tc mi xe.Bi 4: (3, 0 im)Cho ng trn tm O ng knh AB = 2R. Gi C l trung im ca OA, qua C k dy MN vung gcvi OA ti C. Gi K l im ty trn cung nh BM, H l giao im ca AK v MN.a) Chng minh t gic BCHK l t gic ni tip.b) Chng minh AK.AH = R2c) Trn KN ly im I sao cho KI = KM, chng minh NI = KB. HNG DN GII:Bi 1:5a) 2x 5 = 0 2 x 5 = 0 2 x = 5 x =2 y x = 2 5x + 5y = 102y = 20 y = 10b) 5x 3y = 105x 3y = 10 y x = 2x = 8c) thi vo lp 10 mn Ton nm 201293 94. www.VNMATH.com A=5 a 3 3 a +1 a2 + 2 a + 8 5 a 3 + =()( ) (a + 2 + 3 a +1 ) ( a 2) ( a2 + 2 a +8 )a 2a +2 a4 ( a 2) ( a + 2) 5a + 10 a 3 a 6 + 3a 6 a + a 2 a 2 2 a 8 a 2 + 8a 16 ( a 2 8a + 16 ) == = ( a 2 )( a +2 ) ( a 2 )( a +2 ) (a 2 )( a +2 ) ( a 4)2== ( a 4) = 4 a a4 ()( 2 3)2 2d) B = 4 + 2 3 + 7 4 3 = 3 +1 + = 3 +1 + 2 3 = 3 +1+ 2 3 = 3Bi 2:a) Vi m = 1 ( P ) v ( d ) ln lt tr thnh y = x 2 ; y = x 2 .Lc phng trnh honh giao im ca ( P ) v ( d ) l: x 2 = x 2 x 2 + x 2 = 0 ca + b + c = 1 + 1 2 = 0 nn c hai nghim l x1 = 1; x2 = 2 .Vi x1 = 1 y1 = 1Vi x2 = 2 y2 = 4Vy ta giao im ca ( P ) v ( d ) l ( 1; 1) v ( 2; 4 ) .b) Phng trnh honh giao im ca ( P ) v ( d ) l:mx 2 = ( m 2 ) x + m 1 mx 2 ( m 2 ) x m + 1 = 0 ( *) .Vi m 0 th ( *) l phng trnh bc hai n x c = ( m 2 ) 4m ( m + 1) = m 2 4m + 4 + 4m 2 4m = 5m 2 + 4 > 0 vi mi m. Suy ra ( *) lun c hai nghim2phn bit vi mi m. Hay vi mi m 0 ng thng (d) lun ct parabol (P) ti hai im phn bit.Bi 3:i 1h30 = 1,5ht a im :- Quy Nhn l A1,5x100-1,5x- Hai xe gp nhau l CA C B- Bng Sn l BGi vn tc ca xe my l x ( km / h ) . K : x > 0 .Suy ra :Vn tc ca t l x + 20 ( km / h ) .Qung ng BC l : 1,5x ( km )Qung ng AC l : 100 1,5x ( km )100 1,5xThi gian xe my i t A n C l : ( h) x 1,5 xThi gian t my i t B n C l :( h)x + 20 100 1,5 x1,5 xV hai xe khi hnh cng lc, nn ta c phng trnh : = x x + 20Gii pt : thi vo lp 10 mn Ton nm 201294 95. www.VNMATH.com100 1,5 x1,5 x= ( 100 1,5 x ) ( x + 20 ) = 1,5 x 2 100 x + 2000 1,5 x 2 30 x = 1,5 x 2x x + 20 2 3x 70 x 2000 = 02= 35 + 3.2000 = 1225 + 6000 = 7225 > 0 = 7225 = 85 35 + 85Phng trnh c hai nghim phn bit : x1 == 40 (tha mn K) 335 85 50x2 = = (khng tha mn K) 3 3Vy vn tc ca xe my l 40 km / h .Vn tc ca t l 40 + 20 = 60 ( km / h ) .KMBi 4: Ea) T gic BCHK l t gic ni tip.HTa c : IAKB = 900 (gc ni tip chn na ng trn) hay HKB = 900 ; HCB = 900 ( gt ) ACOB T gic BCHK c HKB + HCB = 900 + 900 = 1800 t gic BCHK l t gic ni tip.b) AK . AH = R 2N AC AH RD thy ACH AKB ( g .g ) = AK . AH = AC. AB = 2 R = R 2 AK AB 2c) NI = KBOAM c OA = OM = R ( gt ) OAM cn ti O ( 1)OAM c MC l ng cao ng thi l ng trung tuyn (gt) OAM cn ti M ( 2 )( 1) & ( 2 ) OAM l tam gic u MOA = 600 MON = 1200 MKI = 600 KMI l tam gic cn (KI = KM) c MKI = 600 nn l tam gic u MI = MK ( 3) . 1 1 D thy BMK cn ti B c MBN = MON = 1200 = 600 nn l tam gic u MN = MB ( 4 )22 Gi E l giao im ca AK v MI. NKB = NMB = 600 D thy NKB = MIK KB // MI (v c cp gc v tr so le trong bng nhau) mt MIK = 600 khc AK KB ( cmt ) nn AK MI ti E HME = 900 MHE . HAC = 900 AHC Ta c : HME = 90 MHE ( cmt ) HAC = HME mt khc HAC = KMB (cng chn KB )0 AHC = MHE ( dd ) HME = KMB hay NMI = KMB ( 5 )( 3) , ( 4 ) & ( 5) IMN = KMB ( c.g.c ) NI = KB (pcm) thi vo lp 10 mn Ton nm 2012 95 96. www.VNMATH.com S GIO DC V O TOK THI TUYN SINH VO LP 10 THPT NM HC 2012 NG NAIKha ngy : 29 , 30 / 6 / 2012Mn thi : TON HC CHNH THCThi gian lm bi : 120 pht( ny c 1 trang , 5 cu )Cu 1 : ( 1,5 im ) 1 / Gii phng trnh : 7x2 8x 9 = 0 .3x + 2y =1 2 / Gii h phng trnh : 4x + 5y = 6Cu 2 : ( 2,0 im ) 12 +33 2 2 1 / Rt gn cc biu thc : M = ; N= 32 1 2 / Cho x1 ; x2 l hai nghim ca phng trnh : x2 x 1 = 0 .1 1 Tnh : +.x1 x 2Cu 3 : ( 1,5 im ) Trong mt phng vi h trc ta Oxy cho cc hm s :2y = 3x c th ( P ) ; y = 2x 3 c th l ( d ) ; y = kx + n c th l ( d1 ) vi k v n l nhng s thc . 1 / V th ( P ) . 2 / Tm k v n bit ( d1 ) i qua im T( 1 ; 2 ) v ( d1 ) // ( d ) .Cu 4 : ( 1,5 im )Mt tha t hnh ch nht c chu vi bng 198 m , din tch bng 2430 m2 . Tnh chiu di v chiurng ca tha t hnh ch nht cho .Cu 5 : ( 3,5 im ) Cho hnh vung ABCD . Ly im E thuc cnh BC , vi E khng trng B v E khng trng C . V EFvung gc vi AE , vi F thuc CD . ng thng AF ct ng thng BC ti G . V ng thng a i quaim A v vung gc vi AE , ng thng a ct ng thng DE ti im H . AE CD 1 / Chng minh= . AF DE 2 / Chng minh rng t gic AEGH l t gic ni tip c ng trn . 3 / Gi b l tip tuyn ca ng trn ngoi tip tam gic AHE ti E , bit b ct ng trung trc caon thng EG ti im K . Chng minh rng KG l tip tuyn ca ng trn ngoi tip tam gic AHE .HNG DN GII:Cu 1 : ( 1,5 im ) thi vo lp 10 mn Ton nm 201296 97. www.VNMATH.com 4 79 1 / Gii phng trnh : 7x2 8x 9 = 0 ( x1,2 = ) 73x + 2y =1 2 / Gii h phng trnh : ( x ; y ) = (1 ; 2 )4x + 5y = 6Cu 2 : ( 2,0 im ) 12 +3 2 3 + 3 1 / Rt gn cc biu thc : M ===2+ 3 33 2 ( ) 23 22 1 N= == 2 12 1 2 1 2 / Cho x1 ; x2 l hai nghim ca phng trnh : x2 x 1 = 0 . bc S = =1 ; P = = 1 aa1 1 x1 + x 2 1 Nn : + = = = 1 x1 x 2 x1x 2 1Cu 3 : ( 1,5 im )1 / V th ( P ) .2 / ( d1 ) // ( d ) nn k = 2 ; n 3 v i qua im T( 1 ; 2 ) nn x = 1 ; y = 2 . Ta c phng trnh : 2 =1.2 + n n = 0Cu 4 : ( 1,5 im ) Gi x ( m ) l chiu di tha t hnh ch nht ( 49,5 < x < 99 ) Chiu rng ca tha t hnh ch nht l : 99 x ( m ) Theo bi ta c phng trnh : x ( x 99 ) = 2430 Gii c : x1 = 54 ( nhn ) ; x2 = 45 ( loi ) Vy chiu di tha t hnh ch nht l 54 ( m ) Chiu rng ca tha t hnh ch nht l : 99 54 = 45 ( m )Cu 5 : ( 3,5 im ) 1 / Chng minh t gic AEFD ni tip A = D a 1 1 AEF DCE ( g g )AB AE AF 1 = DC DE 2E AE DC = 1 AF DE IK 2 / Ta c A 2 ph vi A11C Ta c E1 ph vi D1 DF b M A1 = D1 HG A 2 = E1 Suy ra t gic AEFD ni tip ng trn ng knh HE Gi I trung im ca HE I l tm ng trn ngoi tip t gic AEFD cng l ng trn ngoi tipAHE thi vo lp 10 mn Ton nm 2012 97 98. www.VNMATH.com I nm trn ng trung trc EG IE = IG V K nm trn ng trung trc EG KE = KG Suy ra IEK = IGK ( c-c-c ) IGK = IEK = 900 KG IG ti G ca ng trn ngoi tip AHE KG l tip tuyn ca ng trn ngoi tip AHES GIO DC V O TO THI TUYN SINH LP 10 THPTBC GIANGNM HC 2012-2013Mn thi : Ton CHNH THC Thi gian : 120 pht khng k thi gian giao Ngy thi 30 thng 6 nm 2012 www.VNMATH.comCu 1. (2 im) 1 1.Tnh- 22- 1 2 .Xc nh gi tr ca a,bit th hm s y = ax - 1 i qua im M(1;5)Cu 2: (3 im)1 2 a- 3 a + 2 1.Rt gn biu thc: A = ( - ).(+ 1) vi a>0,a 4 a - 2 a- 2 aa- 2 2x - 5 y = 9 2.Gii h pt: 3x + y = 5 3. Chng minh rng pt: x 2 + mx + m - 1 = 0 lun c nghim vi mi gi tr ca m.Gi s x1,x2 l 2 nghim ca pt cho,tm gi tr nh nht ca biu thcB = x 21 + x 2 2 - 4.( x1 + x2 )Cu 3: (1,5 im) Mt t ti i t A n B vi vn tc 40km/h. Sau 2 gi 30 pht th mt t taxi cng xut pht i t An B vi vn tc 60 km/h v n B cng lc vi xe t ti.Tnh di qung ng AB.Cu 4: (3 im) Cho ng trn (O) v mt im A sao cho OA=3R. Qua A k 2 tip tuyn AP v AQ ca ng trn(O),vi P v Q l 2 tip im.Ly M thuc ng trn (O) sao cho PM song song vi AQ.Gi N l giao imth 2 ca ng thng AM v ng trn (O).Tia PN ct ng thng AQ ti K.1.Chng minh APOQ l t gic ni tip.2.Chng minh KA2=KN.KP3.K ng knh QS ca ng trn (O).Chng minh tia NS l tia phn gic ca gc PNM .4. Gi G l giao im ca 2 ng thng AO v PK .Tnh di on thng AG theo bn knh R.Cu 5: (0,5im)Cho a,b,c l 3 s thc khc khng v tho mn: a 2 (b + c ) + b 2 (c + a ) + c 2 ( a + b) + 2abc = 0 2013a+ b 2013 + c 2013 = 111 1Hy tnh gi tr ca biu thc Q = 2013 + 2013 + 2013ab c thi vo lp 10 mn Ton nm 201298 99. www.VNMATH.comHNG DN CHM (tham kho) Cu Ni dungim 1 112+12+1 1- 2=- 2= -2 = 2 + 1- 2 =1 2- 1 ( 2 - 1).( 2 + 1)( 2) 2 - 1)KL: 2Do th hm s y = ax-1 i qua M(1;5) nn ta c a.1-1=5 a=6 1KL: 2 1 a 2( a - 1).( a - 2) 0,5A=(-).( + 1) =a ( a - 2) a ( a - 2)a- 2 a- 2 1 0,5 =().( a - 1 + 1) =. a =1a ( a - 2) aKL: 21 2x - 5 y = 9 2x - 5 y = 9 2x - 5 y = 9 y =- 1 3x + y = 5 15 x + 5 y = 25 17 x = 34 x=2 KL: 3 Xt Pt: x 2 + mx + m - 1 = 0 0,25 = m 2 - 4(m - 1) = m 2 - 4m + 4 = (m - 2) 2 0Vy pt lun c nghim vi mi m x1 + x2 = - m 0,25Theo h thc Viet ta c x1 x2 = m - 1Theo bi B = x 21 + x 2 2 - 4.( x1 + x2 ) = ( x1 + x2 ) 2 - 2 x1 x2 - 4.( x1 + x2 ) = m 2 - 2( m - 1) - 4(- m) = m 2 - 2m + 2 + 4m = m 2 + 2m + 1 + 1 = (m + 1) 2 + 1 10,5Vy minB=1 khi v ch khi m = -1KL: 3Gi di qumg ng AB l x (km) x>0 0,25 xThi gian xe ti i t A n B lh0,2540xThi gian xe Taxi i t A n B l : h0,25 605Do xe ti xut pht trc 2h30pht =nn ta c pt2 0,25xx 5- = 40 60 20,25 3x - 2 x = 300 x = 3000,25 thi vo lp 10 mn Ton nm 201299 100. www.VNMATH.comGi tr x = 300 c tho mn KVy di qung ng AB l 300 km. 4 1Xt t gic APOQ cAPO = 900 (Do AP l tip tuyn ca (O) P)AQO = 900 (Do AQ l tip tuyn ca (O) Q) 0,75 APO + AQO = 1800 ,m hai gc ny l 2 gc i nn t gic APOQ l t gic nitipPSMNI AG O KQ 2Xt AKN v PAK c AKP l gc chungAPN = AMP ( Gc ntcng chn cung NP)0,75M NAK = AMP (so le trong ca PM //AQAK NK AKN ~ PKA (gg) = AK 2 = NK .KP (pcm)PK AK 3K ng knh QS ca ng trn (O)Ta c AQ ^ QS (AQ l tt ca (O) Q)M PM//AQ (gt) nn PM ^ QS0,75ng knh QS ^ PM nn QS i qua im chnh gia ca cung PM nh sd PS = sd SM PNS = SNM (hai gc nt chn 2 cung bng nhau)Hay NS l tia phn gic ca gc PNM 4Chng minh c AQO vung Q, c QG ^ AO(theo Tnh cht 2 tip tuyn ctnhau)Theo h thc lng trong tam gic vung ta c 0,75 OQ 2 R 2 1OQ 2 = OI .OA OI = = = ROA 3R 3 1 8 AI = OA - OI = 3R - R = R 3 3Do KNQ ~ KQP (gg) KQ 2 = KN .KP m AK 2 = NK .KP nn AK=KQVy APQ c cc trung tuyn AI v PK ct nhau G nn G l trng tm22 8 16 AG = AI = . R = R33 3 9 5Ta c:0,25 thi vo lp 10 mn Ton nm 2012 100 101. www.VNMATH.com222a (b + c ) + b (c + a ) + c (a + b ) + 2abc = 0 a 2b + a 2c + b 2 c + b 2 a + c 2 a + c 2b + 2abc = 0 ( a 2b + b 2 a ) + (c 2 a + c 2b) + (2abc + b 2 c + a 2 c ) = 0 ab(a + b) + c 2 (a + b) + c(a + b) 2 = 0 0,25 ( a + b)(ab + c 2 + ac + bc ) = 0 ( a + b).(a + c ).(b + c ) = 0*TH1: nu a+ b=0 a =- b a =- b 111Ta c 2013 ta c Q = 2013 + 2013 + 2013 = 1 a + b + c =1 c =1 20132013 abcCc trng hp cn li xt tng t111Vy Q = 2013 + 2013 + 2013 = 1ab cS GIO DC V O TO K THI TUYN SINH LP 10 THPT TNH YN BINM HC 2012 - 2013 CHNH THC Mn thi: TON Thi gian: 120 pht (khng k thi gian giao ) Kha thi ngy 23/6/2012 ( thi c 01 trang, gm 05 cu)Cu 1: (2,0 im)1. Cho hm s y = x + 3 (1)a. Tnh gi tr ca y khi x = 1b. V th ca hm s (1)2. Gii phng trnh: 4x 7x + 3 = 0Cu 2: (2,0 im) Cho biu thc M = + 1. Tm iu kin ca x biu thc M c ngha. Rt gn biu thc M.2. Tm cc gi tr ca x M > 1Cu 3: (2,0 im) Mt i th m phi khai thc 260 tn than trong mt thi hn nht nh. Trn thc t, mi ngy i u khai thc vt nh mc 3 tn, do h khai thc c 261 tn than v xong trc thi hn mt ngy. Hi theo k hoch mi ngy i th phi khai thc bao nhiu tn than?Cu 4: (3,0 im)Cho na ng trn tm O, ng knh AB = 12 cm. Trn na mt phng b AB cha na ng trn (O) v cc tia tip tuyn Ax, By. M l mt im thuc na ng trn (O), M khng trng vi A v B. AM ct By ti D, BM ct Ax ti C. E l trung im ca on thng BD.1. Chng minh: AC . BD = AB.2. Chng minh: EM l tip tuyn ca na ng trn tm O.3. Ko di EM ct Ax ti F. Xc nh v tr ca im M trn na ng trn tm O sao cho din tch t gic AFEB t gi tr nh nht? Tm gi tr nh nht .Cu 5: (1,0 im) Tnh gi tr ca biu thc T = x + y + z 7 bit:x + y + z = 2 + 4 + 6 + 45 thi vo lp 10 mn Ton nm 2012 101 102. www.VNMATH.com thi vo lp 10 mn Ton nm 2012102 103. www.VNMATH.com thi vo lp 10 mn Ton nm 2012102 104. www.VNMATH.com thi vo lp 10 mn Ton nm 2012102 105. www.VNMATH.com thi vo lp 10 mn Ton nm 2012102