10 bài toán oxy
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10 bài toán OXYTRANSCRIPT
Kha hc PenC N3 (Thy: L Anh Tun_Nguyn Thanh Tng) Hcmai.vn Hocmai.vn Ngi trng chung ca hc tr Vit 1 GI TNG CC BNTNG HP 10 BI TON U TIN TRN FB CA THY www.facebook.com/ThayTungToan Gii Vi ba s1, 1, 1 a b c theo nguyn l Dirichlet tn ti t nht hai s cng du hoc c t nht 1 s bng0. Do vai tr, , a b cbnh ng nn khng mt tnh tng qut gi s : ( 1)( 1) 0 a b > 1 ab a b > +2 2 2 2 abc ac bc c > + Suy ra : 2 2 2 2 2 22 2 2 2 a b c abc a b c ac bc c A + + + > + + + + =Ta c 2 2( ) ( 1) 2( ) 1 2( ) 1 A a b c ab bc ca ab bc ca = + + + + > + + Vy 2 2 22 2( ) 1 a b c abc ab bc ca + + + > + + (*) Khi t (*) v kt hpbt ng thc AM GM ta c: 3 3 32( ) 1 12 2( ) 6 6 1 T ab bc ca ab bc caab bc ca ab bc ca ab bc ca> + + + = + + + + + + + + + + 33 33 2( ).6 .6 1 17 ab bc caab bc ca ab bc ca> + + =+ + + + Vi1 a b c = = =th17 T = . Vy gi tr nh nht ca Tl 17. --------------------------------------------------------------------------------------------------------------------------------------- Gii: iu kin :0 xy > (*) . Ta s ch ra h c nghim0 y =bnghai cch sau :Cch 1:(Dng phng php nh gi)
3 2 3(2) 6( 2 1) 2 x x x xy = + 3 2 36( 1) 2 0 x x xy = < ( v 3 20 xy yxy = > theo (*)) 0 x 8 8 0 y y + > > (4) T (3) v (4) suy ra:0 y = Cch 2: (Dng k thut nhn lin hp v nh gi biu thc khng m) ( ) ( )2 2 33(1) 2 2 2 8 2 0 xy y xy y y + + + + = ( )223 32 2 2 0( 8) 2 8 4yy xy y xy yy y + + + =+ + + +( )( )22312 08 1 3y xy yy ( ( + + = (+ + + ( 0 y = ( v ( )( )22312 08 1 3xy yy+ + >+ + + ,0 xy >) Khi h c dng: 22 2( 6) 12 8 x x x= + = 3 26 12 8 0 x x x + + =3 3 22 6 12 8 x x x x = + ( )33 3 3322 ( 2) 2 21 2x x x x x = = = Vy nghim ca h l:32( ; ) ;01 2xy| |= | \ . -------------------------------------------------------------------------------------------------------------------------------------- Gii: Bi 3 (Nguyn Thanh Tng). Trong mt phng ta Oxy , cho hnh bnh hnhABCD c nh 11 1;2 2A| | |\ .. Mt im(1; 1) M nm trong hnh bnh hnh sao cho MAB MCB =v 0135 BMC = . Tm ta nhD, bit rngD thuc ng trn c phng trnh( ) T :2 22 2 3 0 x y x y + + = . Kha hc PenC N3 (Thy: L Anh Tun_Nguyn Thanh Tng) Hcmai.vn Hocmai.vn Ngi trng chung ca hc tr Vit 3 Dng imEsao choABEMl hnh bnh hnh, khi DCEMcng l hnh bnh hnh Ta c: 1 22 11 1A CC E BECMA E= = = ni tip ng trn0180 BEC BMC + = (1) Mt khc :BEC AMD A = A(c.c.c)BEC AMD = (2) . T (1) v (2) suy ra 0180 AMD BMC + = (*) ng trn ( ) T nhn(1; 1) M lm tm v c bn knh5 R MD = = Ta c 3 102MA =. Theo (*) ta c: 0 0180 45 AMD BMC = =Xt tam gicAMD:
2 2 245 3 10 2 252 . .cos 5 2. . 5.2 2 2 2AD MA MD MAMD AMD = + = + =52AD =Suy raD thuc ng trn tm 11 1;2 2A| | |\ . bn knh 52AD = c phng trnh: 2 22 211 1 2511 18 02 2 2x y x y x y| | | | + = + + = ||\ . \ . Khi ta imD l nghim ca h :
2 22 211 18 02 2 3 0x y x yx y x y + + =+ + = 21xy= = hoc 32xy= = (2;1)(3; 2)DD
.Vy(2;1) Dhoc(3; 2) D . ---------------------------------------------------------------------------------------------------------------------------------------
Gii:iu kin: 1 2 x s s (*) Vi iu kin (*) ta c 2(2) (3 2) 2 1 y x x = 22 13 2xyx =
Xt hm s 2 1( )3 2xf xx= vi | | 1;2 x e . Ta c 21'( ) 0(3 2)f xx= < vi | | 1;2 x e ( ) f xnghch bin trn | | 1;2 ( ) (1) 1 f x f s = hay 21 1 1 y y s s s (2*) Bi 4 (Nguyn Thanh Tng). Gii h phng trnh ( )32 21 2 2 1 2 1 (1)3 2 2 1 0 (2)y x x y x xxy x y+ + + = + =( , ) xy e Kha hc PenC N3 (Thy: L Anh Tun_Nguyn Thanh Tng) Hcmai.vn Hocmai.vn Ngi trng chung ca hc tr Vit 4 Cch 1 (Nguyn Th Duy)Bin i (1)ta c: ( )( )( )21 2 2 1 1 y y y x y x + + + = (3) Do1 y =khng l nghim ca(3)nn 211 1(3) 01 1 2 2yy xy y y y x>+ = > s + + (3*) T (2*) v (3*) suy ra:1 y = , khi 1 x =tha mn h. Vy nghim ca h l ( ; ) (1; 1) xy = .Cch 2 (Nguyn Thanh Tng) Ta c ( )2(1) (1 ) 1 ( 1) 2 2 0 y x y x y y + + + + =
(3)Theo (*) v (2*) ta c: ( )2(1 ) 1 01 21 1 ( 1) 2 2 0y xxy y x y y >s s s s + + + > Khi ( )21(3) (1 ) 1 ( 1) 2 2 01xy x y x y yy= = + + + = = (tha mn h) Vy h c nghim( ; ) (1; 1) xy = . Cch 3 (V c Tng) Ta c 2 2(2) (3 2) 2 1 x y y = 222 13 2yxy=(3) vi 223y =Vi iu kin 1 2 x s s222 11 2 1 13 2yyy s s s s Thay (3) vo (1), ta c: 2 2 2 22 2 2 22 1 2 1 2 1 2 11 2 2 1 2 13 2 3 2 3 2 3 2y y y yy yy y y y| | + + + = | | \ .
2 232 2222 21 4 32 (1 ) ( 1) 03 2 3 21 (4 3)( 1)1 1( 2) (1 ) 03 2 3 2y yy y y yy yy y yy y y y yy y + + + + + = ( + + + + + + =( ( V1 1 y s snn 1 0 y >v ta c 221 72 02 4y y y| | + = + > |\ .,y e nn 222 21 (4 3)( 1)1( 2) (1 ) 03 2 3 2y y yy y y yy y ++ + + + > vi mi1 1 y s s Do 1 0 1 y y += = (tha mn). Thay1 y = vo (3) ta c1 x =(tha mn) Vy h c nghim( ; ) (1; 1) xy = . Ch : V d trn ta cth ch ra 21 y sbng cch phn tch: 22 1 2 1 2 113 2 3 3(3 2) 3 3xyx x= = + s + = vi1 x > .
Kha hc PenC N3 (Thy: L Anh Tun_Nguyn Thanh Tng) Hcmai.vn Hocmai.vn Ngi trng chung ca hc tr Vit 5 Gii t23x ay bz c= = = , khi 2 2 2, , 05 2 4 3 60 (*)xyzx xyz y z> + + + = Ta vit li (*) thnh: 2 2 25 2 . (4 3 60) 0 x x yz y z + + + = (2*) Lc ny quan nim (2*) l phng trnh bc hai vi nx , khi : 2 2 2 2 2 2 2 2' 5(4 3 60) ( 15)( 20) (15 )(20 ) yz y z y z y z A = + = = Mt khc vi iu kin (*) ta c: 2 22 24 60 15 03 60 20 0y yz z < > < > Suy ra 2 2(15 )(20 )5yz y zx =. Do 2 2(15 )(20 )05yz y zx x + > = Khi p dng bt ng thc AM GM (Cauchy) vi hai s dng 215 y ; 220 z ta c: 2 22 2 2(15 ) (20 )(15 )(20 ) 35 ( )25 5 10y zyzyz y z y zx + + + += s =Suy ra 22 260 ( ) 10( ) 2535 ( ) 60 ( 5) 60610 10 10 10y z y zy z y zT x y z y z ( + + + + + = + + s + + = = s = Vi 1 a b c = = =tha mn iu kin bi ton v6 T = . Vy gi tr ln nht ca Tl 6. ---------------------------------------------------------------------------------------------------------------------------------------
Gii: Bc 1: Ta s khai thc phng trnh (1) chi ray x = bng hai cch sau : Cch 1:2 2 225(1) 5 5 55x x x x y yy y + + = + + = + + + 2 25 ( ) 5 ( ) x x y y + + = + + (*) Bi 5 (Nguyn Thanh Tng). Cho, , a b cl cc s thc dng tha mn 2 2 25 12 16 27 60 a abc b c + + + = .Tm gi tr ln nht ca biu thc2 3 T a b c = + + . (Da trn tng t bi ca mt bn hi thy) Bi 6 (Nguyn Thanh Tng). Gii h phng trnh ( )( )( )( )2 22 45 5 5 (1)7 2 5 4 2 (2)x x y yx y y+ + + + = + = + ( , ) xy eKha hc PenC N3 (Thy: L Anh Tun_Nguyn Thanh Tng) Hcmai.vn Hocmai.vn Ngi trng chung ca hc tr Vit 6 Xt hm s 2( ) 5 f t t t = + +22 2 25'( ) 1 05 5 5t tt t tf tt t t++ + = + = > >+ + + ,t esuy ra( ) f tng bin v lin tc trn . Khi (*) ( ) ( ) f x f y = x y = hayy x = (3) Cch 2: ( )( )222 22 222 2222 225 55555 5 ( )5(1)5 5 ( )5 55555y yx xy yx x y y ay yy y x x bx xy yx xx x+ + + = =++ + = + + + + + = + + + + = = ++ + Cng v vi v ( ) av ( ) bta c: 2( ) 0 x y y x + = = (3) Bc 2: Thay (3) vo (2) ta c: ( )( )2 4 2 47 2 5 4 2 7 10 14 5 4 0 x x x x x x + = + + + + = (2*) gii(2*) ta c:Cch 1: Ta c: 4 4 2 2 2 2 2 2 22 2 24 4 4 4 ( 2) (2 ) ( 2 2)( 2 2)7 10 14 ( 2 2) 6( 2 2)x x x x x x x x x xx x x x x x + = + + = + = + + + + = + + + + Nn 2 2 2 2(2*) ( 2 2) 6( 2 2) 5 ( 2 2)( 2 2) 0 x x x x x x x x + + + + + + + = +) t 222 22 2a x xb x x = + += + ( ) , 0 a b > phng trnh c dng:
2 26 5 0 a b ab + = ( 2 )( 3 ) 0 a b a b = 2 a b =hoc3 a b =+) Vi 2 23 9 a b a b = = : 2 2 22 2 9( 2 2) 8 20 16 0 x x x x x x + + = + + = (v nghim). +) Vi 2 22 4 a b a b = = :2 2 25 72 2 4( 2 2) 3 10 6 03x x x x x x x+ + = + + = =Thay vo (3) ta c nghim ca h l: 5 7 5 7 5 7 5 7( ; ) ; , ;3 3 3 3xy | | | |+ + e|| ` || \ . \ . ) Cch 2: (S dng k thut nhn lin hp) (3*)( )2 2 43 10 6 4 8 5 4 0 x x x x + + + + =
( )( ) ( )( )2 2 4 2 4 2 43 10 6 4 8 5 4 4 8 5 4 4 8 5 4 0 x x x x x x x x + + + + + + + ++ + =
( )( )( )2 2 4 4 23 10 6 4 8 5 4 9 64 36 0 x x x x x x + + + + + =
( )( )( )22 2 4 2 23 10 6 4 8 5 4 3 6 (10 ) 0 x x x x x x ( + + + + + = (
( )( )2 2 4 2 23 10 6 4 8 5 4 (3 10 6)(3 10 6) 0 x x x x x x x x + + + + + + + =Kha hc PenC N3 (Thy: L Anh Tun_Nguyn Thanh Tng) Hcmai.vn Hocmai.vn Ngi trng chung ca hc tr Vit 7
( )2 2 4 23 10 6 4 8 5 4 (3 10 6) 0 x x x x x x ( + + + + + + =
23 10 6 0 x x + = (4*) hoc 2 410 2 5 4 0 x x x + + + =(5*) +) Cng (5*)vi (2*) ta c: 28 20 16 0 x x + =(v nghim)+) Ta c (4*)5 73x=.Thay vo (3) ta c nghim ca h l: 5 7 5 7 5 7 5 7( ; ) ; , ;3 3 3 3xy | | | |+ + e|| ` || \ . \ . ) ----------------------------------------------------------------------------------------------------------------------------------- Gii: AJi qua(2;1) Jv(2; 4) D nn cphng trn phng trnh:2 0 x = Khi ta imAl nghim ca h :
2 0 2(2;6)2 10 0 6x xAx y y = = + = = GiEl giao im th hai caBJving trn ngoi tip tam gicABC . Khi :
AmE EnCCpD DqB== EnC CpD AmE DqB + = + hay ECD AmE DqB = +(1) Mt khc: ( )1212EBD sdECDDJB sd AmE sdDqB== + (2) T (1) v (2) suy ra: EBD DJB = hay tam gicDBJcn tiD, suy raDB DJ = (*) . Li c 1 2A A DB DC = = (2*) T (*) & (2*) suy ra:DB DJ DC = =hayD l tm ca ng trn ngoi tip tam gicJBC Suy ra, B Cnm trn ng trn tm(2; 4) D bn knh5 DJ =c phng trnh :2 2( 2) ( 4) 25 x y + + = . Khi ta imBl nghim ca h: Bi 7 (Nguyn Thanh Tng). Trong mt phng ta Oxy , cho tam gicABCngoi tip ng trn tm (2;1) J . Bit ng cao xut pht t nhAca tam gicABCc phng trnh2 10 0 x y + =v (2; 4) D l giao im th hai caAJvi ng trn ngoi tip tam gicABC .Tm ta cc nh ca tam gicABCbitBc honh m vBthuc ng thng c phng trnh7 0 x y + + = . Kha hc PenC N3 (Thy: L Anh Tun_Nguyn Thanh Tng) Hcmai.vn Hocmai.vn Ngi trng chung ca hc tr Vit 8 2 2( 2) ( 4) 257 0x yx y + + = + + =34xy= = hoc 29xy= = ( 3; 4)(2;9)BB DoBc honh m nn ta c( 3; 4) B BCi quaBv vung gc vi ng thng2 10 0 x y + =nn c phng trnh:2 5 0 x y = Khi ta im Cl nghim ca h :
2 2( 2) ( 4) 252 5 0x yx y + + = =34xy= =
hoc 50xy= =( 3; 4)(5;0)(5;0)C BCC
Vy(2;6), ( 3; 4), (5;0) A B C .--------------------------------------------------------------------------------------------------------------------------------------- Gii:Cch 1 (Nguyn Thanh Tng) iu kin: 0 2 x s s Vi iu kin trn khi 2 2(2) 2 ( 1) 3 ( 1) 3( 1) ( 1) 5( 1) x y xy y y x x + + + + = + + +
2 2( 1)(2 3 3) ( 1)( 5) y x x x y + + = + +
2 25 2 3 31 1y x xy x+ + =+ + (*) Xt hm s 22 3 3( )1x xf xx +=+ vi| | 0;2 x e .Ta c: 222 4 6'( )( 1)x xf xx+ =+ ; 21'( ) 0 2 4 6 03xf x x xx= = + = = Khi | || |0;20;2(0) 3 min ( ) 1(1) 1max ( ) 35(2)3xxf f xff xfee= = = = =. Do( ) f xlin tc trn on | | 0;2 , suy ra1 ( ) 3 f x s sT (*) 22224 051 3 3 2 013 2 0y yyy yyy y + >+ s s + s+ + s 1 2 y s sBi 8 (Nguyn Thanh Tng). Gii h phng trnh ( ) ( )2 2 2 2(2 ) 2 1 (1)2 2 3 8 3 2 (2)y x x x xyxy xy x y xy x y = + + = +( , ) xy eKha hc PenC N3 (Thy: L Anh Tun_Nguyn Thanh Tng) Hcmai.vn Hocmai.vn Ngi trng chung ca hc tr Vit 9 Ta c: (1) (2 ) (2 ) 2 (2 ) 2 ( 1) 0 y x y x x y x y x x y y x = + + + =(2*) Vi (2 ) 2 00 201 2( 1) 0y xxx yyy x >s s > s s >(2 ) 2 ( 1) 0 y x x y y x + + > Khi (2*) 0(2 ) 2 ( 1) 02xy x x y y xy= = = = = tha mn(2) . Vy h phng trnh c nghim( ; ) (0;2) xy =Nhn xt:Thc ra bi ton trn tc gi mn s liu v tng t bi ton cho im trong thi i hc khi D2013 va qua Tm gi tr ln nht v gi tr nh nht ca hm s 22 3 3( )1x xf xx +=+ trn on | | 0;2 . Sau khi c bin tu, che y v nh nhiu bng cc s liu i km cng vi mt cch tip cn khc th bi ton clm mi v lt xc i kh nhiu. V vy c rt nhiu nhng bi ton tng nh kh li c sinh ra t nhng tng kh n gin.Cch 2 (T Anh) iu kin: 0 2 x s sTa c ( )2(2 )( 1)(1) 1 0 (2 )( 1) 02y xx xy y xx x = + > >+ Vi 1 2 x s s , bin i phng trnh (2): 2 2 2(1 ) (3 3 2 ) 2 8 2 0 y x y x x x x + + + + =(*) phng trnh c nghim th:
2 2 2 4 3 2(3 3 2 ) 4( 1)( 2 8 2) 4 4 3 58 1 0 x x x x x x x x x A = + + + = + >
2 2(4 8 13)( 3 2) 35 25 0 x x x x x + + + > (2*) M ta c: 21 2 ( 1)( 2) 0 3 2 0 x x x x x s s s + s 2 2(4 8 13)( 3 2) 35 25 0 x x x x x + + + (3*) Vi0 1 x s s , bin i phng trnh (2): 2 2 2(1 ) (3 3 2 ) 2 8 2 0 y x y x x x x + + + + = 2 22 21 1( 5)( 1) ( 1)(2 3 3)2 3 3 5x yy x y x xx x y+ + + + = + + = + + Xt hm 21( )2 3 3xf xx x+= +; 2 22(1 )( 3)'( ) 0(2 3 3)x xf xx x += > + ,| | 0;1 x e . Vy( ) f xng bin trn | | 0;1 . 22 21 1 1(0) 3 3 5 1 25 2 3 3 3y xf y y yy x x+ + = > = + > + s s+ + M theo (3*) ta c2 y >nn suy ra2; 0 y x = = . Vy h phng trnh c nghim( ; ) (0;2) xy = Kha hc PenC N3 (Thy: L Anh Tun_Nguyn Thanh Tng) Hcmai.vn Hocmai.vn Ngi trng chung ca hc tr Vit 10 Gii:
Gii: Gi G,Iln lt l trng tm, tm ng trn ngoi tip tam gicABC . GiNl trung im caMA, khi :
23CK CGGKCN CM= = // MNhay GK // AB . DoIl tm ng trn ngoi tip nnMI AB MI GK (1) GiPl trung im caACv doABC Acn tiAnn: / / / / MP BC MK BCGI MKAG BC GI BC (2) T (1) v (2) , suy raIl trc tm ca tam gicMGK
KI MG hayKI CM .Khi KIc phng trnh: 7 5 7 0 x y + = Suy ra ta imIl nghim ca h 7 5 7 02 4 7 0x yx y+ = + + =72x =v 72y = 7 7;2 2I | | |\ . Gi(4 7;5 ) C t t CM + e , khi :25 252 2R IC IC = = =
2 221 7 257 5 74 42 0 02 2 2t t t t t| | | | + + + = + = = ||\ . \ . hoc 2137t = (loi)(4;0) C Gi(4 7 ;5 ) M m m CM + e , khi Kl trng tm tam gicACMnn 5 77 ; 52 2A m m| | |\ . Ta c( ) ( )2 22 2 21(1; 1)2527 6 5 148 168 47 072 12; 47 237 3774AmIA R m m m mAm
=
= + + = + + = | |
| =
\ .
DoAc ta nguyn nn(1; 1) A 1 5; (0; 4)2 2M B| | |\ . (vMl trung im caAB ) Vy(1; 1), (0; 4), (4;0) A B C . Bi 9 (Nguyn Thanh Tng).Trong mt phng ta Oxy , cho tam gicABCcn tiAvMl trung im caAB . ng thng CMc phng trnh 5 7 20 0 x y =v 11 7;6 6K | | |\ . l trng tm ca tam gicACM . ng trn ngoi tip tam gicABCc tm nm trn ng thng2 4 7 0 x y + + =v c bn knh bng 52. Tm ta cc nh ca tam gicABC , bitAv Cc ta nguyn . Kha hc PenC N3 (Thy: L Anh Tun_Nguyn Thanh Tng) Hcmai.vn Hocmai.vn Ngi trng chung ca hc tr Vit 11
Gii:Ta c 2 2 2 22 4(2 ) 18 2( 2 1) ( 4 4) 8( ) 6 18 b c a b c b b c c a b c + + + + = + + + + + + =
2 224 2( 1) ( 2) 8( ) 8( ) b c a b c a b c = + + + + > + + Suy ra3 a b c ++sDo 0 a b c s s s , nn ta c 2 2 2 2 2 2 2( )( ) ( ) ab bc ca ab bc ca aa b b c abc b a c + + s + + + + = + p dng bt ng thc AM GM dng 3( )27x y zxyz+ +s, ta c: 32 34 2 2( ) 4. . . 4. ( ) 42 2 27 27a c a cba c a cb a c b a b c+ + | |+ + |+ +\ .+ = s = + + sKhi 2 2 24 ab bc ca + + s (*) p dng bt ng thc AM GMdng2xy x y s +v 33xyz x y z s + + , ta c: ( )3 32 5 6 4 2 5 3.2 .1 2.3 2 . .2 a b b bc a b b b c + + = + +2 5 3( 1) 2(2 2) 2( ) 7 a b b b c a b c s + + + ++ = + + + 2.3 7 13 s + = Suy ra ( )32 5 6 4 13 a b b bc + + s Mt khc 0 1 b s s, khi : ta c: ( ) ( )35 6 6 5 0 2 5 6 4 0 b b b b a b b bc + = > + + > Vy ( )30 2 5 6 4 13 a b b bc < + + s (2*) T (*) v (2*), suy ra 134 313P s = Vi 0; 1; 2 a b c = = =tha mn iu kin bi v 3 P = . Vy gi tr ln nht caPbng 3. CM N CC BN C TI LIU ! Nguyn Thanh Tnghttps://www.facebook.com/ThayTungToan Bi 10 (Nguyn Thanh Tng). Cho, , a b cl cc s thc tha mn 0 1 a b c s s s sv 2 22 4(2 ) 18 b c a b c + + + + = . Tm gi tr ln nht ca biu thc: ( )2 2 23132 5 6 4P ab bc caa b b bc= + + + +.