V c Huy - THCS Hoi Thanh Hoi Nhn Bnh nh hS gio dc v o to TP Hi PhngTrng THPT thi tuyn sinh vo lp 10Nm hc: 2008 - 2009 thi ny gm c 01 trangI. Phn trc nghim:Khoanh trn vo ch ci trc cu tr li ng trong cc bi tp sau:Cu 1: ng thng y = ax qua im M(-3 ; 2) v im N(1 ; -1) c phng trnh l:A. y = 4143+ x B. y = -4143 x C. y = 3132 x D. y = 3132+ xCu 2: Phng trnh x4 2mx2 3m2 = 0 ( m 0 ) c s nghim l:A. V nghim B. 2 nghim C. 4 nghim D. khng xc nh cCu 3: Phng trnh 915 x 322xx = x - 3 xx c tng cc nghim l:A. 4 B. - 4 C. -1 D. 1Cu 4:Cho a + 90o. H thc no sau y l SAI ?A. 1- sin2 a = sin2 B. cot ga = tg C. tg = sinD. tga = cotg(90o )Cu 5: Tam gic ABC cn nh A, ng cao AH c AH = BC = 2a. Din tch ton phn ca hnh nn khi cho tam gic quay mt vng xung quanh AH l:A. a2 ( 1 3 + ) B. a2 ( 2 3 + ) B. a2( 1 5 + ) D. a2 ( 2 5 + )Cu 6: cho tga = 43, gi tr ca biu thc C = 5sin2 a + 3cos2 a l:A. 3,92 B. 3,8 C. 3,72 D. 3,36 II Phn t lun:Bi 1: Cho P =

,_

+xxx x11 x

,_

++xxx x11.a. Rt gn Pb. Tm x p < 7 -3 4Bi 2: Cho parabol (P) y = x2 v ng thng (d) y = 2x + m.a. V (P) v (d) trn cng mt h trc to vi m = 3 v tm to giao im ca (P) v (d).b. Tm M (d) tip xc vi (P). Xc nh to tip im.Bi 3: t im M ngoi ng trn (O; R) v tip tuyn MA n ng trn. E l trung im AM; I, H ln lt l hnh chiu ca E v A trn MO. T I v tip tuyn MK vi (O)a. chng minh rng I nm ngoi ng trn (O; R).b. Qua M v ct tuyn MBC ( B nm gia M v C ). Chng minh t gic BHOC ni tipc. Chng minh HA l phn gic ca gc BHC v tam gic MIK cn.http://violet.vn/voduchuy1982/ 1V c Huy - THCS Hoi Thanh Hoi Nhn Bnh nh S gio dc v o to TP Hi PhngTrng THPT p n tuyn sinh vo lp 10Nm hc: 2008 - 2009p n ny c 1 trangI Phn trc nghimCu 1: BCu 2: BPhng trnh trung gian c ac = -3m2 < 0 suy ra phng trnh trung gian c hai nghim tri du uy ra phng trnh c hai nghim.Cu 3: DCu 4: DCu 5: CTa c I = AC =5 asuy ra Stp = RL + R2 = a.a 5 + a2( 1 5 + )Cu 6: CII Phn t lun:Bi 1: a. A = (1- x)2, vi x 0; x 1 b. P < 7- 4 3 1 - x >2 - 33 - 1< x < 3- 3 ; x 1Bi 2: a. Vi m = 3 (d) l y = 2x +3, th i qua im (0; 3) v (0 ;23 )( Bn c t v th)Honh giao im l nghim ca phng trnh x2 = 2x =3Giao im ca parabol v ng thng (d) l (-1 ; 10 ) v ( 3 ; 9 )b. (P) tip xc vi (d) th phng trnh x2 = 2x + m c nghim kp x2 2x m = 0 c = 1 = m = 0 m = -1Bi 3: Bn lm t v hnh.a. Ta c OI2 + IE2 = OE2 = OA2 + EA2 (1)M IE < ME = EA. Vy IE2 < AE2 OI2 > OA2 OI > OA = R (2)T 2 suy ra im I nm ngoi (O; R)b. D dng chng minh c MA2 = MB.MCp dng h thc lng trong tam gic vung AMO, ta c MA2 = MH.MO MBH MOCH1 =C1 t gic BHOC ni tip.c. T trn ta c CHO = B1 =C1 = H1.Vy BHA =AHC( cng ph vi cc gc bng nhau)Ta c HA l phn gic gc BHCIK2 = IO2 R2 (3). T (1) suy ra OI2 + IE2 = R2 = AE2IO2 R2 = AE2 IE2 = ME2 IE2 = MI2 (4)T (3) v (4) suy ra IK = IM, vy tam gic MIK cn ti Ihttp://violet.vn/voduchuy1982/ 2V c Huy - THCS Hoi Thanh Hoi Nhn Bnh nh THI TS VO 10 TNH HI DNGNm hc : 2008 2009Kho thi ngy 26/6/2008 - Thi gian 120 phtCu I: (3 im)1) Gii cc phng trnh sau:a)5.x 45 0 b) x(x + 2) 5 = 02) Cho hm s y = f(x) = 2x2a) Tnh f(-1)b) im( )M 2; 1 c nm trn th hm s khng ? V sao ?Cu II: (2 im)1) Rt gn biu thc P = 4 a 1 a 11 .a a 2 a 2 _ + _

+ , , vi a > 0 v a4.Cu III: (1 im)Tng s cng nhn ca hai i sn xut l 125 ngi. Sau khi iu 13 ngi t i th nht sang i th hai th s cng nhn ca i th nht bng 23 s cngnhn ca i th hai. Tnh s cng nhn ca mi i lc u.Cu IV: (3 im)Cho ng trn tm O. Ly im A ngoi ng trn (O), ng thng AO ct ng trn (O) ti 2 im B, C (AB < AC). Qua A v ng thng khng i qua O ct ng trn (O) ti hai im phn bit D, E (AD < AE). ng thng vung gc vi AB ti A ct ng thng CE ti F.1) Chng minh t gic ABEF ni tip.2) Gi M l giao im th hai ca ng thng FB vi ng trn (O). Chng minh DMAC.3) Chng minh CE.CF + AD.AE = AC2.Cu V: (1 im)Cho biu thc :B = (4x5 + 4x4 5x3 + 5x 2)2 + 2008.Tnh gi tr ca B khi x = 1 2 1.2 2 1+http://violet.vn/voduchuy1982/ 3V c Huy - THCS Hoi Thanh Hoi Nhn Bnh nh GiiCu I:1) a)5.x 45 0 5.x 45 x 45 : 5 x 3. b) x(x + 2) 5 = 0 x2 + 2x 5 = 0 = 1 + 5 = 6' 6 . Phng trnh c hai nghim phn bit : x1,2 =1 6 t .2) a) Ta c f(-1) = 2( 1) 12 2 .b) im( )M 2; 1 c nm trn th hm s y = f(x) = 2x2. V ( )( )22f 2 12 .Cu II:1) Rt gn: P = 4 a 1 a 11 .a a 2 a 2 _ + _

+ , , = ( ) ( ) ( ) ( )( ) ( )a 1 a 2 a 1 a 2a 4.aa 2 a 2 + + += ( ) ( )a 3 a 2 a 3 a 2a 4.a a 4 + + + = 6 a 6a a .2) K: > 0 1 + 2m > 0 m > 12 .Theo bi :( ) ( ) ( )22 2 2 21 2 1 2 1 21 x 1 x 5 1 x x x x 5 + + + + + ( ) ( )221 2 1 2 1 21 x x x x 2x x 5 + + + .Theo Vi-t : x1 + x2 = 2 ; x1.x2 = -2m. 1 + 4m2 + 4 + 4m = 5 4m2 + 4m = 0 4m(m + 1) = 0 m = 0 hoc m = -1.i chiu vi K m = -1 (loi), m = 0 (t/m).Vy m = 0.Cu III:Gi s cng nhn ca i th nht l x (ngi). K: x nguyn, 125 > x > 13.S cng nhn ca i th hai l 125 x (ngi).Sau khi iu 13 ngi sang i th hai th s cng nhn ca i th nht cn li l x 13 (ngi)i th hai khi c s cng nhn l 125 x + 13 = 138 x (ngi).Theo bi ra ta c phng trnh : x 13 = 23(138 x) 3x 39 = 276 2x 5x = 315 x = 63 (tho mn).Vy i th nht c 63 ngi.i th hai c 125 63 = 62 (ngi).Cu IV:http://violet.vn/voduchuy1982/ 4V c Huy - THCS Hoi Thanh Hoi Nhn Bnh nh MMFFEEDDBBCCOOA 3) Xt hai tam gic ACF v ECB c gc C chung , 0A E 90 . Do hai tam gic ACF v ECB ng dng AC ECCE.CF AC.CBCF CB (1).Tng t ABD v AEC ng dng (v c BAD chung, 0C ADB 180 BDE ). AB AEAD.AE AC.ABAD AC (2).T (1) v (2) AD.AE + CE.CF = AC.AB + AC.CB = AC(AB + CB) = AC2.Cu V:Ta c x = ( )( ) ( )22 11 2 1 1 2 12 2 2 2 12 1 2 1 ++ . x2 = 3 2 24 ; x3 = x.x2 = 5 2 78 ; x4 = (x2)2 =17 12 216 ; x5 = x.x4 = 29 2 4132.Xt 4x5 + 4x4 5x3 + 5x 2 = 4. 29 2 4132 + 4. 17 12 216 - 5. 5 2 78 + 5. 2 12 - 2= 29 2 41 34 24 2 25 2 35 20 2 20 168 + + + = -1.Vy B =(4x5 + 4x4 5x3 + 5x 2)2 + 2008 = (-1)2 + 2008 = 1 + 2008 = 2009.http://violet.vn/voduchuy1982/ 51) Ta c 0FAB 90 (V FA AB).0BEC 90 (gc ni tip chn na ng trn (O)) 0BEF 90 0FAB FEB 180 + .Vy t gic ABEF ni tip (v c tng hai gc i bng 1800).2) V t gic ABEF ni tip nn 1AFB AEB2 sAB. Trong ng trn (O) ta c 1AEB BMD2 sBD.Do AFB BMD . M hai gc ny v tr so le trong nn AF // DM. Mt khc AFAC nn DMAC.V c Huy - THCS Hoi Thanh Hoi Nhn Bnh nh S GIO DC V O TO K THI TUYN SINH LP 10 TRNG THPT CHUYNQUNG NAM Nm hc 2008-2009MnTONThi gian lm bi 150 pht ( khng k thi gian giao )Bi 1 ( 1im ):a) Thc hin php tnh:nb3 512 6 3 20 10 3 +.b) Tm gi tr nh nht ca biu thc2008 x x .Bi 2 ( 1,5im ):Cho h phng trnh:' + 5 my x 32 y mxa) Gii h phng trnh khi 2 m .b) Tm gi tr ca m h phng trnh cho c nghim (x; y) tha mn h thc 3 mm1 y x22+ + .Bi 3 (1,5im ):a) Cho hm s 2x21y , c th l (P). Vit phng trnh ng thng i qua hai im M v N nm trn (P) ln lt c honh l2 v 1.b) Gii phng trnh: 1 x x 2 x 3 x 32 2 + +.Bi 4 ( 2im ):Cho hnh thang ABCD (AB // CD), giao im hai ng cho l O. ng thng qua O song song vi AB ct AD v BC ln lt ti M v N.a) Chng minh:1ABMOCDMO + .b) Chng minh:.MN2CD1AB1 +c) Bit 2COD2AOBn S ; m S . Tnh ABCDS theo m v n (viCOD AOBS , S, ABCDS ln lt l din tchtam gic AOB, din tch tam gic COD, din tch t gic ABCD).Bi 5 ( 3im ): Cho ng trn ( O; R ) v dy cung AB c nh khng i qua tm O; C v D l hai im di ng trn cung ln AB sao cho AD v BC lun song song. Gi M l giao im ca AC v BD. Chng minh rng:a) T gic AOMB l t gic ni tip.b) OMBC.c) ng thng d i qua M v song song vi AD lun i qua mt im c nh.Bi 6 ( 1im ):a) Cho cc s thc dng x; y. Chng minh rng: y xxyyx2 2+ + .http://violet.vn/voduchuy1982/ 6V c Huy - THCS Hoi Thanh Hoi Nhn Bnh nh b) Cho n l s t nhin ln hn 1. Chng minh rng n 44 n + l hp s.S GIO DC V O TO K THI TUYN SINH LP 10 TRNG THPT CHUYNQUNG NAM Nm hc 2008-2009MnTON Thi gian lm bi 150 pht ( khng k thi gian giao )HNG DN CHM MN TON I. Hng dn chung:1) Nu th sinh lm bi khng theo cch nu trong p n m vn ng th cho im tng phn nh hng dn quy nh.2) Vic chi tit ha thang im (nu c) so vi thang im trong hng dn chm phi m bo khng sai lch vi hng dn chm v c thng nht trong Hi ng chm thi.3) im ton bi ly im l n 0,25.II. p n:Bi Ni dung im1(1)a) Bin i c:2 2 33 5) 2 2 3 )( 3 5 (+ + 0,250,25b) iu kin2008 x 4803148031)212008 x (412008 )412008 x .21. 2 2008 x ( 2008 x x2 + + + Du = xy ra khi48033x212008 x (tha mn). Vy gi tr nh nht cn tm l 48033x khi48031 .0,250,252(1,5)a) Khi m = 2 ta c h phng trnh ' + 5 y 2 x 32 y x 2' +' + 2 x 2 y55 2 2x5 y 2 x 32 2 y 2 x 2'+56 2 5y55 2 2x0,250,250,25b) Gii tm c: 3 m6 m 5y ;3 m5 m 2x2 2+++Thay vo h thc 3 mm1 y x22+ + ;ta c3 mm13 m6 m 53 m5 m 2222 2+ ++++Gii tm c 74m 0,250,250,25http://violet.vn/voduchuy1982/ 7 CHNH THCV c Huy - THCS Hoi Thanh Hoi Nhn Bnh nh 3(1,5)a) Tm c M(- 2; - 2); N )21: 1 ( Phng trnh ng thng c dng y = ax + b, ng thng i qua M v N nn' + + 21b a2 b a 2Tm c1 b ;21a . Vy phng trnh ng thng cn tm l1 x21y 0,250,250,25b) Bin i phng trnh cho thnh 0 1 x x 2 ) x x ( 32 2 + +t x x t2+ ( iu kin t 0 ), ta c phng trnh0 1 t 2 t 32 Gii tm c t = 1 hoc t = 31 (loi)Vi t =1, tac0 1 x x 1 x x2 2 + +. Gii rac25 1x+ hoc 25 1x .0,250,250,254(2)Hnh v OABC DNM0,25a) Chng minh c ADMDABMO;ADAMCDMO Suy ra1ADADADMD AMABMOCDMO + + (1)0,250,50b) Tng t cu a) ta c1ABNOCDNO + (2)(1) v (2) suy ra2ABMNCDMNhay 2ABNO MOCDNO MO + +++Suy ra MN2AB1CD1 +0,250,25c)n . m S n . m SSSSSOCOAODOB;OCOASS;ODOBSSAOD2 2 2AODCODAODAODAOBCODAODAODAOB Tng t n . m SBOC . Vy 2 2 2ABCD) n m ( mn 2 n m S + + + 0,250,25Hnh v(phc v cu a)0,25http://violet.vn/voduchuy1982/ 8V c Huy - THCS Hoi Thanh Hoi Nhn Bnh nh 5(3) OICDMBAa) Chng minh c: - hai cung AB v CD bng nhau -s gc AMB bng s cung AB Suy ra c hai gc AOB v AMB bng nhau O v M cng pha vi AB. Do t gic AOMB ni tip 0,250,250,250,25b) Chng minh c: - O nm trn ng trung trc ca BC (1) - M nm trn ng trung trc ca BC (2)T (1) v (2) suy ra OM l ng trung trc ca BC, suy raBC OM 0,250,250,25c) T gi thit suy raOM d Gi I l giao im ca ng thng d vi ng trn ngoi tip t gic AOMB, suy ra gc OMI bng 090 , do OI l ng knh ca ng trn nyKhi C v D di ng tha mn bi th A, O, B c nh, nn ng trn ngoi tip t gic AOMB c nh, suy ra I c nh. Vy d lun i qua im I c nh.0,250,250,250,256(1)a) Vi x v y u dng, ta cy xxyyx2 2+ + (1) 0 ) y x )( y x ( ) y x ( xy y x2 3 3 + + + (2)(2) lun ng vi mi x > 0, y > 0. Vy (1) lun ng vi mi 0 y , 0 x > >0,250,25b) n l s t nhin ln hn 1 nn n c dng n = 2k hoc n = 2k + 1, vi k l s t nhin ln hn 0.- Vi n = 2k, ta c k 2 4 n 44 ) k 2 ( 4 n + +ln hn 2 v chia ht cho 2. Do n 44 n +l hp s.-Vi n = 2k+1, tac

2 k 2 k 2 2 k 4 k 2 4 n 4) 2 . n . 2 ( ) 4 . 2 n ( ) 4 . 2 ( n 4 . 4 n 4 n + + + + = (n2 + 22k+1 + n.2k+1)(n2 + 22k+1 n.2k+1) = [( n+2k)2 + 22k ][(n 2k)2 + 22k ]. Mi tha s u ln hn hoc bng 2. Vy n4 + 4n l hp s0,250,25======================= Ht ======================= thi v li giihttp://violet.vn/voduchuy1982/ 9V c Huy - THCS Hoi Thanh Hoi Nhn Bnh nh http://violet.vn/voduchuy1982/ 10V c Huy - THCS Hoi Thanh Hoi Nhn Bnh nh Li gii mn Ton Bi I.Cho biu thc x xxxxxP+

,_

++ :11a) Rt gn P( ) ( )( ) ( )( )xx xPxx xx xx x xx xPx xxx xx xx xxxxxP11 .1111:111:11:11+ ++++ ++ ++ ++ ++ ++

,_

++ b) Tnh gi tr ca P khi x = 4Vi x = 4 th 2741 4 4+ + Pc) Tm x 313 Pkx: x>0( ) 0 3 10 3 13 1 3313 1313 + + + + + x x x x xxx xP(1)tt x ; iu kin t > 0 Phng trnh (1)0 3 10 32 + t t ; Gii phng trnh ta c

313tt (tho mn iu kin)*) Vi t = 39 3 x x*) Vi 913131 x x tBi II. Gii bi ton bng cch lp phng trnhGi s chi tit my t th nht lm c trong thng u l x (xN*; x < 900; n v:chi tit my)S chi tit my t th hai lm c trong thng u l 900-x (chi tit my)http://violet.vn/voduchuy1982/ 11V c Huy - THCS Hoi Thanh Hoi Nhn Bnh nh Thng th hai t I lm vt mc 15% so vi thng th nht nn t I lm c 115%x=1,15x (chi tit my)Thng th hai t II lm vt mc 10% so vi thng th nht nn t II lm c 110%(900-x)=1,1(900-x) (chi tit my)Thng th hai c hai t lm c 1010 chi tit my nn ta c phng trnh:1,15x + 1,1(900-x)= 1010 1,15x + 1,1.900 1,1.x= 10100,05x = 20 x = 20:0,05 x = 400 (tho mn iu kin) vy thng th nht t I sn xut c 400 chi tit myt II sn xut c 900 400 = 500 chi tit my.Bi III. Cho Parabol (P) 241x y v ng thng (d) y = mx + 11) Chng minh vi mi gi tr ca m ng thng (d) lun ct parabol(P) ti hai im phn bit.Xt phng trnh honh giao im ca (d) v (P):(*) 0 4 4 1412 2 + mx x mx xHc sinh c th gii theo mt trong hai cch sau:Cch 1. m m m > + + 0 4 4 4 ) 2 ( '2 2 (*) lun c hai nghim phn bit vi mi gi tr ca m (d) lun ct (P) ti hai im phn bit vi mi gi tr ca m.Cch 2. V a.c = 1. (-4) = -4 a=-1;a=-4http://violet.vn/voduchuy1982/ 58OKFEDCBAV c Huy - THCS Hoi Thanh Hoi Nhn Bnh nh T ta c' + + 12 3 24 ) ( 3 ) (2y xy x y x

* ' + 12 3 21y xy x(1)*' + 12 3 24y xy x(2)Gii h (1) ta c x=3, y=2Gii h (2) ta c x=0, y=4 Vy h phng trnh c nghim l x=3, y=2 hoc x=0; y=4b)Ta c x3-4x2-2x-15=(x-5)(x2+x+3)m x2+x+3=(x+1/2)2+11/4>0 vi mi x Vy bt phng trnh tng ng vi x-5>0 =>x>5Cu 3: Phng trnh: ( 2m-1)x2-2mx+1=0 Xt 2m-1=0=> m=1/2 pt tr thnh x+1=0=> x=1 Xt 2m-1 0=> m1/2 khi ta c, = m2-2m+1= (m-1)2 0 mi m=> pt c nghim vi mi mta thy nghim x=1 khng thuc (-1,0)vi m1/2 pt cn c nghim x=1 21+ mm m=1 21 m

pt c nghim trong khong (-1,0)=> -1m E,F thuc ng trn ng knh BKhay 4 im E,F,B,K thuc ng trn ng knh BK.b.BCF=BAF M BAF=BAE=450=> BCF= 450Ta cBKF= BEFM BEF= BEA=450(EA l ng cho ca hnh vung ABED)=>BKF=450V BKC= BCK= 450=> tam gic BCK vung cn ti B 2Bi 1: Cho biu thc: P = ( )

,_

+

,_

++11 2 2:1 1xx xx xx xx xx xa,Rt gn Phttp://violet.vn/voduchuy1982/ 59V c Huy - THCS Hoi Thanh Hoi Nhn Bnh nh b,Tm x nguyn P c gi tr nguyn.Bi 2:Cho phng trnh: x2-( 2m + 1)x + m2 + m - 6= 0 (*)a.Tm m phng trnh (*) c 2 nghim m.b.Tm m phng trnh (*) c 2 nghim x1; x2tho mn 3231x x =50Bi 3: Cho phng trnh: ax2 + bx + c = 0 c hai nghim dng phn bit x1, x2Chng minh:a,Phng trnh ct2 + bt + a =0 cng c hai nghim dng phn bit t1 v t2.b,Chng minh: x1 + x2 + t1 + t24Bi 4: Cho tam gic c cc gc nhn ABCni tip ng trn tm O . H l trc tm ca tam gic. D l mt im trn cung BC khng cha im A.a, Xc nh v tr ca im D t gic BHCD l hnh bnh hnh.b, Gi P v Q ln lt l cc im i xng ca im D qua cc ng thng AB v AC . Chng minh rng 3 im P; H; Q thng hng.c, Tm v tr ca im D PQ c di ln nht.Bi 5: Cho hai s dng x; y tho mn: x + y 1Tm gi tr nh nht ca: A = xy y x501 12 2++p nBi 1: (2 im). K: x 1 ; 0 x a, Rt gn: P = ( )( )( )11 2:11 22xxx xx x zP = 11) 1 (12+xxxx b. P= 12111+ +x xx P nguyn th) ( 1 2 19 3 2 10 0 1 14 2 1 1Loai x xx x xx x xx x x

Vy vi x={ } 9 ; 4 ; 0th P c gi tr nguyn.http://violet.vn/voduchuy1982/ 60V c Huy - THCS Hoi Thanh Hoi Nhn Bnh nh Bi 2: phng trnh c hai nghim m th:( ) ( )'< + +> + + + 0 1 20 60 6 4 1 22 122 12 2m x xm m x xm m m

3210 ) 3 )( 2 (0 25 < ' + > mmm m b. Gii phng trnh: ( ) 50 ) 3 ( 23 3 + m m

' + + + + 25 125 10 1 50 ) 7 3 3 ( 5212 2mmm m m m

Bi 3: a. V x1 l nghim caphng trnh: ax2 + bx+ c = 0 nn ax12 + bx1 + c =0. .V x1> 0 =>c. . 01.1121 + + ,_

axbx Chng t 11x l mt nghim dng ca ph-ng trnh: ct2 + bt + a = 0; t1 = 11xV x2 l nghim ca phng trnh: ax2 + bx + c = 0=> ax22 + bx2 + c =0v x2> 0 nn c.01.1222 +

,_

+

,_

axbx iu ny chng t 21x l mt nghim dng ca phng trnh ct2 + bt + a = 0 ; t2 = 21x

Vy nu phng trnh: ax2 + bx + c =0 c hai nghim dng phn bit x1; x2 th phng trnh :ct2 + bt + a=0cng c hai nghim dng phn bit t1 ; t2 . t1 = 11x ; t2 =21xb. Do x1; x1; t1; t2 u l nhng nghim dng nn t1+ x1 = 11x + x12 t2 + x2 = 21x + x2 2 Do x1 + x2 + t1+ t2 4 http://violet.vn/voduchuy1982/ 61V c Huy - THCS Hoi Thanh Hoi Nhn Bnh nh Bi 4a. Gi s tm c im D trn cung BC sao cho t gic BHCD l hnh bnh hnh . Khi : BD//HC; CD//HB v H l trc tm tam gic ABC nn CHAB v BH AC => BD AB v CD AC . Do : ABD = 900 v ACD = 900 . Vy AD l ng knh ca ng trn tm O Ngc li nu D l u ng knh AD ca ng trn tm O th t gic BHCD l hnh bnh hnh.b) V P i xng vi D qua AB nn APB = ADBnhng ADB =ACB nhngADB =ACBDo :APB =ACBMt khc:AHB + ACB = 1800 =>APB +AHB = 1800

T gic APBH ni tip c ng trn nn PAB =PHBMPAB =DAB do :PHB =DABChng minh tng t ta c:CHQ =DAC VyPHQ =PHB +BHC + CHQ =BAC +BHC = 1800Ba im P; H; Q thng hng c). Ta thy APQ l tam gic cn nh A C AP = AQ = AD vPAQ =2BAC khng i nn cnh y PQ t gi tr ln nht AP v AQ l ln nht hay AD l ln nht D l u ng knh k t A ca ng trn tm O Lun chc mi ngi hn phc v lun vui vhttp://violet.vn/voduchuy1982/ 62HOPQDC BA