WWW.VIETMATHS.COMCHUYN 1:Phng trnh v h phng trnh.I.Gii phng trnh bng cch t n ph thch hp.Bi 1:Gpt:2 2222 2 410. 11. 0.1 1 1x x xx x x + _ _ _+ + , , ,Gii:t 2 2;1 1x xu vx x + + (1).Ta c: 10.u2 + v2 -11.uv = 0(u-v).(10u-v)=0u=v hoc 10u=v.Xt cc trng hp thay vo (1) ta tm c x mt cch d dng.Bi 2:Gpt: (x2 - 4x+3).(x2 - 6x + 8)=15.Gii:t x2 - 5x + 5 = u (1).Ta c: (x2 - 4x+3).(x2 - 6x + 8)=15(x-1).(x-3).(x-2).(x-4)-15=0(x-1).(x-2).(x-3).(x-4)-15=0(x2-5x+4).(x2-5x+6)-15=0(u-1).(u+1)-15=0u2-16=0u= t4.Thay cc gi tr ca u vo (1) ta d dng tm c x.Bi 3:Gpt:290.1 1x xx x _ _+ + , ,Gii:PT22 21 1. 90( 1) ( 1)xx x 1+ 1+ ].222 22 2. 90( 1)xxx+ .t u = x2( u0)(1).Ta c:2 222 2. 90 2 2 90.( 1)( 1)uu u u uu+ + ( u 1).0 90 182 882 + u u.T y ta d dng tm c u, thay vo (1) ta tm c x.Bi 4:Gpt:3 332 3 12.( 1) x x x + .Gii:t 3 3; 2 3 x u x v (1). 1WWW.VIETMATHS.COMC: ) .( 4 ) .( 3 ) .( 43 3 3 333 3v u v u uv v u v u v u + + + + + +

+ + + v uv uv u v u v u v u v u 0 ) ) . ( . ( 3 0 ) 2 ) . ( . ( 32 2 2Xt cc trng hp thay vo (1) ta d dng tm c x.Bi 5:Gpt: xxx x x 32 212 3 3 522 3+ + + +(1).Gii:T (1) suy ra:1 6 2 3 3 5 . 22 2 3 + + + x x x x xx x x x x x x x 12 2 12 1 36 8 12 12 202 3 2 4 2 3 + + + + + 0 9 24 22 82 3 4 + + x x x x(x 0).09 2422 822 + + x xx x.t yxx +3 (*) ta c:y2 - 8y + 16 = 0 suy ra y = 4 thay vo (*) ta d dng tm c x.Bi 6:Gpt: ( ) ). 1 ( 0 1841). 4 .( 3 ) 4 .( 1 + + +xxx x xGii: iu kin x > 4 hoc x < -1.*Nu x > 4, (1) tr thnh:0 18 ) 4 ).( 1 ( . 3 ) 4 ).( 1 ( + + + x x x xt 0 ) 4 ).( 1 ( + y x x (2) ta c:y2 + 3y -18 = 0.T ta d dng tm c y,thay vo (2) ta tm c x.*Nu x < -1, (1) tr thnh:0 18 ) 4 ).( 1 ( . 3 ) 4 ).( 1 ( + + x x x xt 0 ) 4 ).( 1 ( + y x x (3) ta c:y2 - 3y -18 = 0.T ta d dng tm c y,thay vo (3) ta tm c x.Bi 7:Gpt:(2x2 - 3x +1).(2x2 + 5x + 1)=9x2 (1).Gii:(1) 0 1 2 20 4 42 3 4 + + + x x x x(x 0).Chia c hai v cho x2 ta c :4x2 + 4x -20 + 21 2x x + = 0. 0 2412 . 2122 ,_

+ + ,_

+xxxx . t y = xx12 +.(2)Ta c: y2 + 2y -24 = 0.T ta tm c y,thay vo (2) ta d dng tm c x.Bi 8:Gpt: . 0 16 8 . 2 64 162 2 2 + + + x x x x xGii:PT. 0 4 . 2 8 + x x x 2x - 048+x-8---0+x-4-- 0 + +x -0++ +WWW.VIETMATHS.COMn y ta xt tng khong ,bi ton tr nn n gin.Bi 9:Gpt: (1 + x + x2)2 = 5.(1 + x2 + x4).Gii:4 2 3 2 4 25 5 5 2 2 2 1 x x x x x x x + + + + + + + 4 3 2 4 3 24 2 2 2 4 0 2 2 0 x x x x x x x x + + + + Nhn thy x = 0 khng phi l nghim ca phng trnh cho, vy x 0.Chia c hai v ca phng trnh trn cho x2 ta c:2x2 - x + 1 - 02 12 +x x. t y = xx1+ (*). Ta c:2y2 - y - 3 = 0.T ta d dng tm c y, thay vo (*) ta tm c x.Bi 10: Gpt: (6-x)4 + (8-x)4 = 16.Gii:t 7 - x = y (*).Ta c: (y-1)4 + (y + 1)4 =162y4 +12 y2 +2 = 162.(y-1).(y+1).(y2+7)=0y =1 hoc y = -1.Thay cc gi tr ca y tm c trn thay vo (*) ta d dng tm c cc gi tr ca x.II.Tm cc nghim nguyn (x;y) hoc (x;y;z) ca cc phng trnh sau:Bi 1: x2 = y.(y+1).(y+2).(y+3)Gii:t y2 + 3y = t.Ta c: x2 = y.(y+1).(y+2).(y+3) = (y2 + 3y).(y2 + 3y +2) = t2 + 2t.*Nu t > 0 th t2 < x2 = t2 + 2t < (t+1)2 suy ra khng tn ti x tha mn.*Nu t < -2 th 2t + 4 < 0nn t2 + 2t > t2 + 4t + 4 suy ra t2 + 2t > t2 + 4t + 4 = (t+2)2.Suy ra: x2 = t2 + 2t > (t + 2)2 (*).Li c: t2 +2t < t2 suy ra x2 < t2 (**).T (*)&(**) suy ra (t + 2)2 < x2 < t2 suy ra x2 = (t+1)2 suy ra t2 +2t = (t +1)2 (=x2)Suy ra : t2 +2t = t2 +2t +1 (V l).*Nu t = -1 suy ra x2 = t2 +2t = -1 0 v . 0 , >xyyxtA=. 3 + +yzxxyzzxyGi s z 0.Ta c:A = 33 . 3 . . . . . 3 . . 3 z xyxyzyxzzxyyxzxyzzxyyzxxyzzxy + + + +

1 , 11 , 11 , 1 . 1y x zy x zx y z z x yBi 6: 2x2 - 2xy = 5x + y - 19.Gii:T bi ra ta c: . 17 , 1 1 2 1 2 171 21721 219 5 22t t + + ++ + ++ + x xxxx x xy T ta tm c x tm c y.III.Gii h phng trnh v cc phng trnh khc.Bi 1:. 221 12+xxGii:iu kin :2 , 0 < x x.-Nu x < 0 th 0:t x = a vb x 22 (a,b > 0). 4WWW.VIETMATHS.COMTa c:' + +221 12 2b ab aC:11. 21 12 + abab b a (1).Li c: 2 = a2 + b22ab suy ra 1 ab (2).T (1)&(2) suy ra ab = 1 m a2 + b2 =2 nn suy ra (a+b)2 = 4 suy ra a + b = 2.Vy ta c:1 121 ' +x b ab aa b.Bi 2: . 5 16 3 2 4 1 44 2 2 2+ + + + + y x y y x x xGii:iu kin:' + + ) 4 ( 0 1 6) 3 ( 0 3 2) 2 ( 0 4 1) 1 ( 0 442 22xy y xxxT (4) suy ra x24 kt hp vi (1) suy ra x2 = 4 kt hp vi (2) suy ra x = 2.Phng trnh cho tr thnh:5 1 + y y .Lc ny vic tm y khng cn kh khn g na (Lp bng xt du).Bi 3: 2x4 -21x3 + 74x2 -105x +50 =0.Gii:Nhn thy x = 0 khng phi l nghim ca phng trnh cho.Vy x 0.Chia c hai v ca phng trnh cho cho x2 ta c:0 2625. 2125. 2 050 10574 21 2222 ,_

+ ,_

+ + + xxxxx xx xt yxx +25 ta c:2y2 -21.y - 26 = 0.T ta tm c ytm ra x.Bi 4:' + + +7 1 . 4 15 1 1 . 2x xx x 5WWW.VIETMATHS.COMGii:t :' + 0 10 1x bx aH cho tr thnh:' + 7 45 2b ab aT tm c a =3,b =1.n y vic tm ra x khng cn kh khn na.Bi 5:' + + ) 2 ( 1 5) 1 ( 1 5 1x yy xGii:Thay biu thc (2) vo phng trnh (1) ta c:1 1 . 2 1 5 1 5 1 + + x x x.T ta tm c x.Vic tm gi tr ca y cng khng c g kh khan na.Bi 6:' + + + + ) 2 ( 0 3 3 2) 1 ( 0 2 4 4 5 1 2 4 1 5 22 22 2x y x y y xy x y x y xGii:Phng trnh (2) phn tch c nh sau:(x - y).(x -3 + 2y) = 0

y xy x2 3Xt cc trng hp thay vo phng trnh (1) ta d dng tm c x v y.Bi 7: x3 + (3-m).x2 + (m-9).x + m2 -6m + 5 = 0.Gii:Phng trnh cho phn tch c nh sau:[ ] [ ] 0 ) 1 ( 2 . ) 5 (2 m x x m x .n y vic gii v bin lun phng trnh khng cn kh khn g na. 6WWW.VIETMATHS.COMBi 8:' + + + +x y z z y xz y x4 4 41Gii:B :. : , ,2 2 2ca bc ab c b a R c b a + + + + ng thc xy ra khi v ch khi a = b = c. (D dng chng minh c b trn).S dng b ta c:xyz = x4 + y4 + z4x2y2 + y2z2 + z2x2xyz.(x + y + z) = xyz.Suy ra cc du bt ng thc trn u phi tr thnh ng thc tc l ta phi c:x = y =zkt hp vi gi thit ban u :x + y + z =1 ta c:31 z y x.Bi 9:( ) '+ + + +) 2 ) ( 2 0 0 1 . () 1 ( 12 0 0 02 0 0 0 1 9 9 91 9 9 92 2x y y x x y y x y xGii:iu kin: x,y. 0 Nhn nhn phng trnh (2) ta thy:-Nu x > y th: VT > 0, VP < 0 suy ra: VT > VP.-Nu y > x th:VT 0 suy ra: VT < VP.-Nu x = y khi : VT =VP =0.Kt hp vi (1) (Ch :x,y . 0 ) ta c: 21 y x.Bi 10: 2 . 2 2 5 2 . 3 2 5 2 + + + x x x x(1).Gii:(1)( ) ( ) 2 . 2 3 3 2 .211 5 2 .212 2 + + x x 4 3 5 2 1 5 2 + + x xTa c:. 4 1 5 2 5 2 3 1 5 2 5 2 3 4 + + + + x x x xVy du bt ng thc trn phi tr thnh du ng thc tc l:2575 2 90 5 20 5 2 3 ' xxxxVy nghim ca phng trnh cho l:1]1

7 ;25x. 7WWW.VIETMATHS.COMCHUYN 2:Bt ng thc.Cc bi ton tm gi tr ln nht , nh nht.Bi 1:Cho a,b,c l di ca ba cnh tam gic.CMR: ab + bc + ca a2 +b2 +c2 < 2.(ab + bc + ca).Gii:Ta c:a2 +b2 +c2 - ab + bc + ca[ ] . 0 ) ( ) ( ) ( .212 2 2 + + a c c b b ang thc xy ra khi v ch khi a = b = c.Vy: ab + bc + ca a2 +b2 +c2.Li c:a < b + c a2 < a.(b + c) (1)Tng t: b2 < b.(a + c) (2) ,c2 < c.(b + a) (3).Cng (1),(2),(3) theo v ta c:a2 +b2 +c2 < a.(b + c) + b.(a + c) + c.(b + a) = 2.(ab + bc + ca).Bi 2:Gi s x > z ; y > z ; z > 0.CMR:xy z y z z x z + ) .( ) .((1).Gii:t:'+ + n z ym z x (m,n,z > 0).Khi (1) tr thnh: ) ).( ( n z m z zn zm + + +( ) z nzmn m + ,_

+ + . 1 (2).p dng BT Bunhiacopxki ta c:( )221 .( ) . 1 .( ) .m m mn z n z n m n z n mz z z _ _ _+ + + + + + + , , ,Vy (2) ng, tc l (1) cng ng (pcm).Bi 3:Cho xy > 0 v x + y = 1.CMR:( ) . 51. 84 4 + +xyy xGii:T gi thit . 0 ,0 10> '> +>y xy xx yTa c:). 1 ( 4141. 2 1 + xyxy xy y xLi c:( ) ( )2224 4 2 2 4 4 2 2 2 2 2 2 28. 4.(1 1 ).( ) 4.( ) (1 1 ).( ) 1. x y x y x y x y x y 1 1 + + + + + + + ] ] 8WWW.VIETMATHS.COMSuy ra:8.(x4 + y4) 1 (2).T (1) v (2) suy ra:( ) . 5 4 11. 84 4 + + +xyy xTa c pcm.Bi 4:Cho ba s phn bit a,b,c.CMR:C t nht mt trong ba s sau y l s dng:x = (a + b + c)2 - 9ab ; y = (a + b + c)2 - 9cb ; z = (a + b + c)2 - 9ac.Gii:Ta c:x + y + z = 3. (a + b + c)2 - 9.(ab + bc + ca) = 3.(a2 + b2 +c2- ab - bc - ca) = = [ ] . 0 ) ( ) ( ) ( .232 2 2> + + a c c b b a (Do a b c a).Vy trong ba s x,y,z lun c t nht mt s dng.Bi 5: Nu'> +01a bb a th 814 4 +b a.Gii: Hon ton tng t bi 3.Bi 6:CMR: ( ) ( ) ( ) ( )4 4 8 8 2 2 10 10. . y x y x y x y x + + + + .Gii:Ta c:( ) ( ) ( ) ( )4 4 8 8 2 2 10 10. . y x y x y x y x + + + +( ) ( )4 4 4 4 12 12 8 8 2 2 12 12. . y x y x y x y x y x y x + + + + + + ( ) ( )4 4 4 4 8 8 2 2. . y x y x y x y x + + ( ) 0 .6 2 2 6 8 8 2 2 + y x y x y x y x( ) ( ) ( ) ( )22 2 2 2 6 6 2 2 2 2 4 2 2 4. . 0 . . 0 x y x y x y x y x y x x y y + + Bt ng thc cui cng lun ng.Vy ta c pcm.Bi 7:CMR: Nu a,b,c l cc s i mt khc nhau v a + b + c < 0 th :P = a3 + b3 + c3 - 3abc < 0.Gii:C:P = a3 + b3 + c3 - 3abc = (a + b + c).(a2 + b2 + c2 - ab - bc - ca) < 0.Bi 8:CMR:41) 1 2 (1...251912 n nGii:D dng bin i tng ng chng minh c:1]1

+ +++1.Gii:Ta c: k k k k k111). 1 (1 12 y vxy = 1.CMR:. 0 2 22 2 +y xy xGii:Ta c:. 0 2 22). ( . 222 2 + +y xy xy xy xy xy xTa c pcm.Bi 12:Cho tam gic ABC c cc cnh tha mn:. c b a CMR: ( ) . 92bc c b a + +Gii:T gi thit bi ra ta c:( ) ) 1 ( 9 2 5 40 ) 4 ).( ( 0 4 22 2 2bc c b bc c bc b c b c b c a b b + + > > + M: (a + b + c)2(2b + c)2 (2).T (1) v (2) suy ra:(a + b + c)2(2b + c)29bc.Ta c pcm.Bi 13: Cho 0 < a,b,c < 2.CMR:Ba s a.(2-b) ; b.(2-c) ; c.(2-a) khng ng thi ln hn 1.Gii:Ta c:. 122.22.22) 2 ( ). 2 .( ). 2 .( ) 2 ( ). 2 ( ). 2 .(2 2 2 ,_

+

,_

+

,_

+ c c b b a ac c b b a a a c c b b aTch ca ba s nh hn hoc bng 1 v vy chng khng th ng thi ln hn 1.Ta c pcm.Bi 14: Cho ba s a,b,c tha mn: a > b > c > 0.CMR: 10WWW.VIETMATHS.COMc a c acb a b ab +< +.Gii:Ta c: c a c acb a b ab +< +2 2 2 2 2 2 2 2 2 22 22 2. 2 2.a b a b a c a ca b a b a c a ca a b a a c a b a c b c+ + + + < + + < + + + < + < >Bt ng thc cui cng lun ng.Vy ta c pcm.Bi 15:Cho cc s dng x,y,z tha mn:. 12 2 2 + + z y xCMR:. 13 3 3 + +xzzyyxGii:p dng BT C Si:23 32 . 2 x xyyxxyyx +(1).Tng t:232y yzzy + (2)v232z xzxz + (3).Cng (1),(2),(3) theo v ta c:) .( 22 2 23 3 3z y x zxxzyzzyxyyx+ + + + + + +Suy ra:. 1 ) ( ) ( ) .( 22 2 2 2 2 23 3 3 + + + + + + + + z y x zx yz xy z y xxzzyyxVy ta c pcm.CHUYN 3: a thc v nhng vn lin quan. Bi 1:Cho 1 2 2&2 352 32+ ++ +x xbx aQx x xP . Vi nhng gi tr no ca a,b th P=Q vi mi gi tr ca x trong tp xc nh ca chng.Gii:iu kin:. 1 , 2 xTa c: P=Q 1 , 22 32 ) 2 (2 35) 1 , 2 (3232 + + + + xx xb a x b a axx x xx' ' +215 20 21bab ab aaBi 2:Cho s nguyn n, A= n5 - n.a-Phn tch A thnh nhn t. 11WWW.VIETMATHS.COMb-Tm n A=0.c-CMR: A chia ht cho 30.Gii:a) A= n5 - n = n.(n4 -1) = n.(n-1).(n+1).(n2 + 1)b) A=0n = 0,1,-1.c) Theo nh L Fecma: 5 5 ) 5 (mod5 5 A n n n n (1).Li c:2 2 ) 1 ( A n n (2) v:3 3 ) 1 .( ). 1 ( A n n n + (3).V 2,3,5 i mt nguyn t cng nhau nn t (1),(2)&(3) suy ra ) 5 . 3 . 2 ( A (pcm).Bi 3: CMR: Nu x,y l nhng s nguyn tha mn iu kin x2 + y2 chia ht cho 3 th c x v y u chia ht cho 3.Gii:Nhn xt:S chnh phng chia cho 3 c s d l 0 hoc 1.V vy t gi thit x2 + y2 chia ht cho 3. 3 , y x Bi 4:Tm gi tr ca p,q a thc (x4 + 1) chia ht cho a thc x2 + px + q.Gii:Gi s (x4 + 1) = (x2 + px + q).( x2 + mx + n)Khai trin v ng nht h s ta c h:'+ ' + + +qq pq np mq nq p m np m111002Vy c th thy cc gi tr ca p,q cn tm l:'+ > qq pq10Bi 5:Cho a thc:120 154 71 14 ) (2 3 4+ + x x x x x A Z x .a)Phn tch A(x) thnh nhn t.b)Chng minh a thc A(x) chia ht 24.Gii:a).Ta c: 120 154 71 14 ) (2 3 4+ + x x x x x A3 2 2( 2).( 12 47 60) ( 2).( 3).( 9 20) x x x x x x x x + +b).Ta c:A(x)= 242) (120 144 72 ) 14 ).( 1 ).( 1 ( + + + x x x x x xx B-Nu x chia ht cho 4,x-14 chia ht cho 2 B(x) chia ht cho 8. 12WWW.VIETMATHS.COM-Nu x chia cho 4 d 1 th x-1 chia ht cho 4,x+1 chia ht cho 2B(x) chia ht cho 8.-Nu x chia cho 4 d 2 th x-14 chia ht cho 4,x chia ht cho 2B(x) chia ht cho 8.-Nu x chia cho 4 d 3 th x + 1 chia ht cho 4,x-1 chia ht cho 2B(x) chia ht cho 8.Vy trong mi trng hp ta u c B(x) chia ht cho 8 (1).M tch ca ba s nguyn lin tip th chi ht cho 3 nn (x-1).x.(x+1) chia ht cho 3B(x) chia ht cho 3 (2).M (3,8)=1 nn t (1) v (2) suy ra B(x) chia ht cho 24.Vy ta c pcm.Bi 6:Tm tt c cc s nguyn x : x2 + 7 chia ht cho x-2.Gii:Ta c: x2 + 7 = (x-2).(x + 2) +11 chia ht cho x-2 khi v ch khi 11 chia ht cho x-2. x-2=-1,-11,1,11.T ta d dng tm ra cc gi tr x tha mn bi ra.Bi 7:Mt a thc chia cho x-2 th d 5, chia cho x-3 th d 7.Tnh phn d ca php chia a thc cho (x-2).(x-3).Gii:Gi a thc cho l F(x).Theo bi ra ta gi s a thc d cn tm l ax+b.Ta c:F(x) = (x-2).(x-3).A(x) + ax + b. (trong A(x) l a thc thng trong php chia)Theo gi thit v theo nh l Bdu ta c:F(2)=2a +b=5 v F(3)=3a+b=7.Gii h hai phng trnh trn ta tm c a = 2, b = 1.Vy a thc d l 2x+1.Bi 8: Cho bit tng cc s nguyn a1, a2, a3..., an chia ht cho 3.Chng minh rng:A(x) = 3 3231...na a a + + + cng chia ht cho 3.Gii:Theo nh l fecma ta c:Z n n n ) 3 (mod3.p dng ta c: ) 3 (mod131a a ,) 3 (mod232a a ,...,) 3 (mod3n na a .Suy ra: 3 3231...na a a + + + ) 3 (mod 0 ) 3 (mod ...2 1 + + + na a aTa c pcm.Bi 9:Chng minh rng (7.5n 2+12.6n) lun chia ht cho 19, vi mi s n t nhin.Gii:Ta c:A =7.52n + 12.6n = 7.25n + 12.6n.Ta c: ) 19 (mod 6 25 ) 19 (mod 6 25n n .Suy ra:) 19 (mod 0 ) 19 (mod 6 . 19 6 . 12 6 . 7 + n n nA.Ta c pcm. Bi 10: Phn tch thnh nhn t x10 + x5 + 1.Gii:Ta c: x10 + x5 + 1 = (x2 + x + 1).(x8-x7 + x5-x4 + x3-x + 1).CHUYN 4: Cc bi ton lin quan ti phng trnhbc hai v nh l Vi-et. 13WWW.VIETMATHS.COMBi 1:Cho phng trnh : x2 -(2m+1)x + m2+m -1= 01.Chng minh phng trnh lun c nghim vi mi m.2.Chng minh c mt h thc gia hai nghim s khng ph thuc vo m.Gii:1. Ta c := (2m +1)2 - 4.(m2 + m - 1) = 5 > 0suy ra phng trnh lun c nghim vi mi m2.Theo vi-et ta c:' + + +) 2 ( 1 .) 1 ( 1 222 12 1m m x xm x xT (1) suy ra: 212 1 +x xmthay vo (2) ta c: ,_

++ ,_

+ 12121.2 122 12 1x x x xx x 12121.2 122 12 1 ,_

+ ,_

+x x x xx x.Ta c pcm.Bi 2: Tm nhng gi tr nguyn ca k bit thc ca phng trnh sau l s chnh phng: k.x2 + (2.k-1).x + k-2= 0; (k 0)Gii:Ta c := (2k-1)2 - 4.k.(k-2) =4k +1 .Gi s 4k + 1 l s cp khi n l s cp l hay: 4k + 1 = (2n + 1)2 n l s t nhin.Hay: k = n2 + n.Vy l s cp th k = n2 + n( th li thy ng).Bi 3: Tm k phng trnh sau y c ba nghim phn bit :(x-2)(x2 + k.x + k2 - 3)= 0Gii:t f(x)= (x-2)(x2 + k.x + k2 - 3) = (x-2).g(x) f(s)=0 c ba nghim phn bit tng ng vi g(x) =0 c hai nhgim phn bit khc 2 hay:' > >'> 12 20 ) 2 (0 ) 3 . ( 42 2kkgk kBi 4: Tm a,b hai phng trnh sau l tng ng: x2 + (3a + 2b) x - 4 =0 (1) v x2 + (2a +3b)x + 2b=0 (2)vi a v b tm c hy gii cc phng trnh cho.Gii:-iu kin cn: 14WWW.VIETMATHS.COMNhn thy pt (1) lun c 2 nghim phn bit.Vy pt (2) cng phi c 2 nghim phn bit ging vi (1).t f(x) = x2 + (3a + 2b) x - 4 =0 v g(x) = x2 + (2a +3b)x + 2b. hai phng trnh cho l tng ng th f(x) = g(x) (*) vi mi x (V h s ca x2 ca c hai pt u bng 1).Thay x = 0 vo (*) ta c b = -2 (3).Thay x = 1 vo (*) kt hp vi (3) ta c a= -2.-iu kin :Vi a=b=-2 ta thy hai phng trnh tng ng vi nhau.Bi 5: Gi s b v c l cc nghim ca phng trnh : x2 - a.x-21.a2 =0;(a 0) chng minh : b4 + c4 2+ 2.Gii:Theo nh l Viet ta c:' +221ab ca c bTa c: [ ]2 222 2 2 2 2 2 4 42 2 ) ( 2 ) ( c b bc c b c b c b c b + + +2 2 2 6 223. . 2 22321 144444222 4 4+ > + + + + + ,_

+ + aaaaa aa c b .Bi 6 : Chng minh rng vi mi a,b,c phng trnh sau lun c nghim : a(x-b).(x-c) + b.(x-c). (x-a) + c.(x-a).(x-b) = 0Gii: t f(x) = a.(x-b).(x-c) + b.(x-c). (x-a) + c.(x-a).(x-b) = = (a + b + c).x2 -2.(ab + bc + ca).x + 3abc*Nu a + b + c 0.Khi :' = a2b2 + b2c2 + c2a2 -abc.(a + b + c) = [(ab-bc)2 + (bc-ca)2 + (ca-ab)2].210 *Nu a + b + c = 0.Khi :-Nu ab + bc + ca0 th phng trnh cho lun c nghim.-Nuab+bc+ca=0. Khi kt hpvi gt a+b+c=0taddngchngminhc a=b=c=0.V d nhin trng hp ny pt cho c v s nghim.Bi 7:CMR:Nu cc h s a,b,c ca phng trnh:ax2+ bx + c = 0 (a 0) u l cc s l th phng trnh bc hai trn khng th c nghim hu t.Gii:Gi s phng trnh bc hai trn vi cc h s a,b,c u l cc s l c nghim hu tx0 = nmvi m,n l cc s nguyn (m,n)=1 v n 0 ;khi ta c:a. 0 .2 + + ,_

cnmbnm hay: 02 2 + + cn bmn am(1).Suy ra: 15WWW.VIETMATHS.COM'n a mm c n22 m (m,n)=11 ) , ( ) , (2 2 n m m n nn:'n a m c m c,a u l cc s l nn suy ra m,n cng l cc s l.Vy ta c:a,bc,m,n u l cc s l .Do : + +2 2cn bmn am s l (Mu thun vi (1)).Vy iu ta gi s l sai.Hay ni cch khc, ta c pcm.CHUYN 5: Cc bi ton hnh hc phng mang yu t chuyn ng.Bi 1: Cho ng trn (O) v dy cung BC c nh.Gi A l im di ng trn cung ln BC ca ng trn (O), (A khc B,C).Tia phn gic ca gc ACB ct ng trn (O) ti im D khc C, ly im I thuc on CD sao cho DI = DB.ng thng Bi ct ng trong (O) ti im K khc im B.1.CMR:Tam gic KAC cn.2.CMR: ng thng AI lun i qua im c nh J.T tm v tr ca A sao cho Ai c di ln nht.3.Trn tia i AB ly im M sao cho AM=AC.Tm tp hp cc im M khi A di ng trn cung ln BC ca (O).Gii:1.Ta c:DBI cn ti D nn: DBI= DIB.M:DIB =IBC +ICB (1).V:DBI =KCI =KCA +ACD =KBA +ICB (2).T (1) v (2) suy raABI =CBI.Suy ra I l tm ng trn ni tip tam gic ABC BI l phn gic gc B ca tam gic ABCK l trung im cung AC. Tam gic KAC cn.2.V I l tm ng trn ni tip tam gic ABC nn AI lun i qua trung im J ca cung nh BC.Ta d dng chng minh c tam gic BIJ cn JJI = JB = const.Suy ra AI = AJ - IJ = AJ - const ln nht khi v ch khi AJ ln nht tc l AJ l ng knh ca (O) A phi nm ti trung im ca cung ln BC.3.Ta d dng tnh c:BMC = 21. BAC =41s o cung nh BC = const.Suy ra qu tch im M l cung cha gc nhn BC di mt gc bng 41s o cung nh BC.Bi 2:Trn ng trn tm O bn knh R ly im A c nh v im B thay i.ng vung gc vi AB v t A ct ng trn C.1. Chngminh rng BC i qua mt im c nh.2.Gi AH l ng vung gc v t A ca tam gic ABC.Tm tp hp cc im H3. Hy dng tam gic vung ABC c nh A cho trc trn ng trn BC l ng knh v chiu cao AH = h cho trc.Gii:1.D thy BC lun i qua im O c nh.2.Nhn thyAHO vung. T d dng chng minh c qu tch ca H l ng trn ng knh AO. 16WWW.VIETMATHS.COM3.ng thng d // vi BC cch BC mt khong h ct (O) ti hai im A v A' tha mn yu cu ca bi ton.C 4 v tr ca A tha mn bi ra (V c hai ng thng d//BC tho mn:Cch BC mt khong h).Bi 3:Cho ng trn tm O c nh .Mt ng thng d c nh ct (O) ti A,B;M l im chuyn ng trn d ( ngoi on AB).T M k hai tip tuyn MT v MN vi ng trn.1.CMR:ng trn i qua ba im M,N,P lun i qua mt im c nh khc O.2.Tm tp hp cc tm I ca ng trn i qua M,N,P.3.Tm trn d mt im M sao cho tam gic MNP l tam gic u.Gii:1.Gi K l trung im ca AB.D thy M,N,P,O,K u nm trn ng trn ng knh OM.Vy K l im c nh cn tm.2. Tm I ca ng trn i qua M,N,P l trung im ca OM.T I h IJ vung gc vi AB.D thy IJ = (1/2).OK=const.Vy c th phn on qu tch ca i l ng thng song song vi AB cch AB mt khong bng mt na on OK tr on XY vi X,Y ln lt l trung im ca OA v OB.3.Gi s tam gic MNP u th th: OM = 2.OP = 2R.MK2 = MO2 - OK2 = 4R2 - OK2 = const.T c hai im M tho mn bi ra.Bi 4:Cho hnh vung EFGH.Mt gc vung xEy quay xung quanh im E.ng thng Ex ct ng thng FG v GH ti M,N;cn ng thng Ey ct cc ng trn theo th t ti P,Q.1.CMR:Hai tam gic ENP v EMQ l cc tam gic vung cn.2.Goi R l giao ca PN v QM;cn I,K ln lt l trung im ca PN v QM.T gic EKRI l hnh g?Gii thch?3.CMR: F,K,H,I thng hng.T c nhn xt g v ng thng IK khi gc vung xEy quay quanh E?Gii:1.D dng chng minh c:EHQ =EFM (cgc).Suy ra d dng tam gic EMQ vung cn.PEF =PQN (ng v) mFEM =QEH.Suy ra:PEN =PEF +FEM =EQH +QEH = 900.Vy tam gic PEN vung (1).Thy:NEQ =PEM (gcg) nn suy ra EN = EP (2).T (1) v (2) suy ra:Tam gic PEN vung cn.2.C: EI PN vEKQM.Vy t gic EKRI c gc I v gc K vung (4).Li c:PQR =RPQ = 450 suy ra:PRQ = 900 (3).T (3) v (4) suy ra t gic IK l hnh ch nht.3.D thy QEKH v EFMK l cc t gic ni tip.Ta c:EKH = 1800 -EQH (5).V:EKF =EMF = EQH (6).T (5) v (6) suy ra:EKH +EKF = 1800. Suy ra H,K,F thng hng. 17WWW.VIETMATHS.COMLi c:T gic FEPI ni tip nnEFI = 1800- EPI = 1800-450 = 1350.Suy ra:EFK + EFI = 450 + 1350 =1800.Suy ra K,F,I thng hng.Vy ta c pcm.Bi 5:Cho ng trn tm O ng knh AB.Gi C l im c nh trn OA; M l im di ng trn ng trn.Qua M k ng vung gc vi MC ct cc tip tuyn k t A v B D v E.a)CMR: Tam gic DCE vung.b)CMR: Tch AD.BE khng i.c)CMR:Khi M chy trn ng trn th trung im I ca DE chy trn mt ng thng c nh.Gii:a)Nhn thy cc t gic ADMC v MABE l cc t gic ni tip.Do :DCM =DAM vMCE =MBE =MAB.Vy:DCE =DCM +MCE =DAM +MAB = 900.Ta c pcm.b)V tam gic DCE vung C nn ta c th nhn thy ngayDCA = 900 - ECB = CEB.Vy hai tam gic vung ADC v BCE ng dng vi nhau.Nn:. . . const AC BC BE ADBEACBCAD c)Nhn thy OI lun l ng trung bnh ca hnh thang DABE hay ni cch khc,ta lun c OI AB.Vy khi M chuyn ng trn (O) th I lun nm trn ng thng qua O vung gc vi AB.Bi 6:Cho tam gic ABC cn (AB=AC) ni tip ng trn tm O.M l im bt k chy trn y BC.Qua M v ng trn tm D tip xc vi AB ti B.V ng trn tm E qua M tip xc vi AC ti C.Gi N l giao im th hai ca hai ng trn .. CMRa) MN lun i qua A v tch AM.AN khng i.c) Tng hai bn knh ca hai ng trn tm D v E c gi tr khng i.d)Tm tp hp cc trung im H ca DE.Gii:a) Ta c: gc BNM = gc ABC =gc ACB =gc BNA.vy tia NM i qua A. Chng minh tam gic ABNng dng vi tam gic AMB suy ra AM.AN = AB2 khng i c)Gi K l im chnh gia ca cung BC ( khng cha A).D thy D,E ln lt nm trn BK v CK. T K,D,E ln lt h cc ng vung gc vi BC ti I.J,L. Ta c:1 1 1. . . 12 2 21BD CE BJ CL BM CM BM CMBK CK BI CI BI CI BIBD CEBD CE CKCK CK++ + + + + =khong oid) H HQ vung gc vi BC.C:HQ = 1.( ) . .2 2 2 2KIDJ EL KI BD CE KIDJ ELKI BK CK+ _+ + ,. Nn H nm trn ng thng song song vi BC cch BC mt khong bng na khong cch KI , V D , E thuc BK v CK do 18WWW.VIETMATHS.COMqu tch cc im H l ng trung bnh ca tam gic BKC (song song vi y BC).CHUYN 6: Cc bi ton hnh hc phng c ni dung chng minh, tnh ton.Bi 1: Cho tam gic OAB cn nh O v ng trn tm O c bn knh R thay i (R OA(trng hp ngc li hon ton tng t).Ta c: |OA2 - OM2| = OM2 -OA2 = MI2 - IA2 = (MI-IA).(MI + IA) = AM.(MT + TB)==MA.MB (pcm).Bi 2:Cho im P nm ngai ng trn (O); Mt ct tuyn qua P ct (O) A v B.Cc tip tuyn k t A v B ct nhau M. Dng MH vung gc vi OP.a)CMR: 5 im O,A,B,M,H nm trn 1 ng trn.b)CMR: H c nh khi ct tuyn PAB quay quanh P. T suy ra tp hp im M.c)Gi I l trung im ca AB v N l giao im ca PA vi MH.CMR: PA.PB=PI.PNv IP.IN=IA2.Gii:a) Nhn thy 5 im O,A,B,M,H nm trnng trn ng knh OM (pcm).b)Phng tch ca im P i vi ng trn ng knh OM l:PH.PO=PA.PB=const (1). Suy ra H c nh nm trn on PO.T d dng suy ra c rng qu tch im M l ng thng d qua H vung gc vi PO tr i on TV vi T,V l giao im ca d vi (O).c)Phng tch ca im P i vi ng trn ng knh ON l: PN.PI=PH.PO (2)T (2) v (1) suy ra: PA.PB=PI.PN (pcm).Li c:IP.IN=(NI+NP).IN=IN2 + NI.NP (3)Phng tch ca im N i vi ng trn ng knh PM l: NP.NI=NH.NMPhng tch ca im N i vi ng trn ng knhOM l: NH.NM=NA.NBSuy ra: NI.NP=NA.NB (4)T (3) v (4) suy ra:IP.IN=IN2 + NA.NB 19WWW.VIETMATHS.COMTa s chng minh: IN2 + NA.NB=IA2 (5).Tht vy:(5)NA.NB=IA2-IN2 NA.NB=(IA-IN).(IA+IN) NA.NB=NA.(IB+IN)NA.NB=NA.NB (lun ng)Vy ta c pcm.Bi 3:Cho tam gic ABC c ba gc nhn ni tip trong ng trn bn knh R,tm O.a)Chng minh BC = 2R.SinAb)Chng minh:SinA + SinB + SinC < 2.(cosA + cosB + cosC) trong A,B,C l ba gc ca tam gic.Gii:a)Ko di BO ct (O) ti im th hai l D.Tam gic vung BCD c:BC = BD.Sin( BDC) = 2R.SinA (pcm)b)Ko di AO ct (O) ti im th hai l E. Hon ton tng t phn a) ta c:AC=2R.SinB. Ta c:SinB=CosA CosC CDB Cos ADB CosBDCDBDADRCD ADRAC+ + + +< ) ( ) (2 2 (1)Tng t ta cng c: SinC < CosA + CosB (2) v SinA < CosB + CosC (3).Cng (1),(2),(3) theo v ta c pcm.

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