Sáng tạo và giải phương trình, bất phương trình, hệ phương trình, bất đẳng thức - Tài liệu ôn thi Đại học môn Toán: Phần 1

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Phần 1 tài liệu Tài liệu ôn thi Đại học môn Toán - Sáng tạo và giải phương trình, bất phương trình, hệ phương trình, bất đẳng thức trình bày các nội dung: Phương pháp giải phương trình, bất phương trình vô tỷ, phương pháp giải hệ phương trình. Mời các bạn cùng tham khảo.

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Nội dung Text: Sáng tạo và giải phương trình, bất phương trình, hệ phương trình, bất đẳng thức - Tài liệu ôn thi Đại học môn Toán: Phần 1

510.76 NGUYEN TRUNG KI£N T103L TAI LIEU ONTHIDAI HOC /cirxg b o o v o gioi PHl/ONG TRlNH X BAT PHl/ONG TRlNH H£ PHl/ONG TRINH X BIT DANG THtfC • D^nh cho hQC sinh Idp 12 chiicfng t r i n h chuan n^ng cao • On tap nang cao k i nang 1km hki m Bien so^n theo npi dung vk cau true di t h i cua B Q G D & D T +6x^-2x + 3-(5x-l)Vx^+3=0 V-4x^+18x-20 + ^"'-^^^SV^ 2x2-9x + 8 4 1 DVL.013496 Chung minh: — + — > 5 (1) x 4y NHA XUATBAN DAI HOC QUOC 6IA HA NOI
TAI LIEU O N T H I D A I H O C Joqg boo VQ gioi X PHl/ONG T R I N H BAT PHl/ONG T R I N H ' ; HE PHUWNG TRINH BAT D A N G THUt • Danh cho hoc sinh Idp 12 chiicrng t r i n h chuan va nang cao • O n tap va nang cao k i nang lam bai • Bien soan theo noi dung va cau true de t h i ciia Bo GD8fDT - I THi; VIEW T I f y ' H B I N H THUAN' OVL '4 / -y. • • • -' 0[}{] t
Ldi noi dHu Phan 1: P h u o n g t r i n h , bat p h u o n g t r i n h , h f p h u o n g t r i n h , bat d 5 n g thuc la m ^ n g PHL/ONG PHAP GlAl PHLTONG TRINH, kien thuc quan t r o n g t r o n g c h u o n g t r i n h toan pho t h o n g . Dac bi^t cac bai toan ve p h u o n g t r i n h , bat p h u o n g t r i n h , h ^ p h u o n g t r i n h , bat d a n g thuc t h u o n g BAT PHLfONG TRlNH VO TY xuyen xua't hien t r o n g cac k y t h i chQn h p c s i n h gioi, cOng n h u T u y e n sinh dai hQC va l u o n gay k h o k h a n cho hoc sinh. I. NHOTNG KI^N T H L f C B6 T R O C H O GlAl P H l / O N G T R I N H V O i t N h S m g i i i p cac e m hpc sinh T H P T cung n h u cac e m hpc sinh chuyen Toan CO m p t tai l i f u m a n g t i n h h f t h o n g de o n l u y f n , nang cao k i e n thuc k y nang 1. G i a i p h u a n g trinh bac 4: giai toan de d a t ket qua cao nha't trong cac k y t h i H g c sinh g i o i , k y t h i T u y e n a) PhuoiTg trinh dang: x^ = ax^ + bx + c sinh dai hpc, c u n g n h u t h i vao cac l o p chuyen chpn, toi bien soan cuon: "Tai P h u o n g phap: Ta t h e m bot vao 2 v e m p t l u p n g : 2mx^ + m ^ k h i d o p h u o n g li^u on thi Dai hoc - sang tao va giai phuong trinh, bai phuong trinh, h$ t r i n h t r o t h a n h : (x^ + m)^ = (2m + a)x^ + bx + c + phuong trinh, bai dang thitc". Ta m o n g m u o n ve'phai c6 dang: ( A x + B)^ N p i d u n g cuon sach d u p e chia l a m 4 phan: 2m + a > 0 ,Phan 1 : P h u o n g phap giai p h u o n g t r i n h , bat p h u o n g t r i n h v 6 t y Phan 2: P h u o n g phap giai h^ p h u o n g t r i n h A = h^- 4(2m + a)(c + m^) = 0 ^ Phan 3: P h u a n g phap h a m so'trong cac bai toan chua t h a m so' Vi 1: Giai cac phuang trinh: Phan 4: P h u o n g phap h a m so trong c h i i n g m i n h bat dang thuc va t i m a) x+V 5 W ^ =6 b)2x2-6x-l =V i ; ^ GTLN, G T N N T r o n g m o i p h a n toi l u o n c6' gang h^ t h o n g p h u o n g phap, p h a n tich, d j n h Giii: h u o n g each giai, cuoi m o i p h a n deu c6 bai tap ren l u y ^ n de cac e m hpc sinh a) D i e u k i ^ n : 1 < x < 6 ' ' thusuc. Dat y = >0. T o i h y v p n g cuo'n sach se la tai li^u b o ich cho cac e m hpc sinh hpc tot m o n P h u o n g t r i n h t r o thanh: y2 + ^/^^^ ^ 5 ^ j y ' " l O y ' - y + 20 = 0 Toan va dat ket qua cao t r o n g cac k y t h i . (0 < y < Vs > ^f,, Mac d i i da co gang d a n h nhieu t a m huyet cho vi?c bien soan cuon sach song thie'u sot la d i e u k h o n g the tranh k h o i . Rat m o n g s u d o n g g o p phe b i n h Xet p h u o n g t r i n h : y4 - I0y2 - y + 20 = 0 o y" = 10y2 + y - 20. ciia ban dpc de Ian tai ban sau d u p e hoan thi^n h o n . Ta t h e m vao 2 v e ' p h u o n g t r i n h m p t l u p n g : 2my^ + m^ C u o i cung toi x i n g u i l a i cam o n sau sac den ban be, d o n g n g h i f p, cac dien K h i d o p h u o n g t r i n h t r o thanh: dan toan da cung cap m p t so'tai l i ^ u quy gia de hoan thi^n cuon sach. y'^ + 2my2 + m^ = (10 + 2m)y^ + y + m^ - 20 T a c gia T a c o Ayp = l - 4 ( m 2 - 2 0 ) ( 1 0 + 2m) = 0 « > m = - - 2 Nguyen Trung Kien Ta Viet lai p h u o n g t r i n h thanh: y ^ - 9 y 2 + (-] 2 1 =y +y+7«> =0 4 y-2
Tdt lieu ou tin h,u si'iii^ tao vd ^fiat fi, oai fi, ne yrrmn^i -nguyen irung IVICTI -i + Vi7 (TM) => II-V17 a) Dieu kifn: x > — « ( y 2 - y - 5 ) ( y 2 + y - 4 ) = 0=*y = X = 4 Binh phuong hai ve'ta thu duoc phuong trinh: b) Dieu ki^n: x > — o x ^ - 6 x ^ +8x2 + 2 x - l = 0 < » x ^ - 6 x ^ = - 8 x 2 - 2 x + l Dat y = V4x + 5 > 0 thi phuong trinh da cho c6 dang: Ta tao ra vetrai dang: (x^ - 3x + m)^ = - 6x^ + (9 + 2m)x2 - 6mx + m^ y ' ' - 2 2 y 2 - 8 y + 77 = 0 » y"* = 22y2 + 8 y - 7 7 Tuc la them vao hai ve mpt lup-ng la:(9 + 2m)x2 - 6 m x + m2 phuong trinh Ta them vao 2 ve phuang trinh mot luong: 2my^ + m^ tro thanh: (x^ - 3x + m)^ = (2m + l)x2 - (6m + 2)x + m^ +1 Khi do phuang trinh tro thanh: Ta can A\p = ( 3 m + 1 ) - ( 2 m + I ) ( m 2 + 1 ) = 0 o m = 0 y^ + 2my^ + m^ = (22 + 2m)y^ + 8y + - 77 Phuang trinh tro thanh: (x^ - 3x)2 = (x -1)^ Ta CO Ayp = 1 - 4(22 + 2m)(m2 - 77) = 0 o m = - 9 . x = 2 + V3 Ta viet lai phuong trinh thanh: x = 2-V3
GiAi: + ( ^ - b ) [ ^ +^b + ^ = a-b^ a) Phuong trinh tuong duang voi: + ( ^ / I - b ) ( V ^ + b) = a-b2 x . - l f / 2 2 x = l i-, + Neu h(x) = 0 C O nghiem x = XQ thi ta luon phan tich dugc +2x + 6 = (2x + l ) 2 3x2+2x-5 = 0 , ... h(x) = (x-Xo)g(x) b) Dieu ki^n: x > 0. Binh phuong 2 ve ta dugc: x>-8 Nhu vay sau buoc phan tich va riit nhan tu chung x - XQ thi phuong trinh 3x +1 + l^lx^ +x = 4x + 9 l^fhJTx =X +8 o 4(2x2+x) = (x +8)2 X - Xg = 0 ban dau tro thanh: (x - Xo)A(x) = 0 A(x) = 0 x>-8 x=4 16 Vi?c con lai la diing ham so', bat dang thuc hoac nhirng danh gia co ban de i 7x2-12x-64 = 0 x = —7 ke't luan A(x) = 0 v6 nghiem. Doi chieu vai dieu kien ta tha'y chi c6 x = 4 la nghi^m cua phuong trinh. • Neu phuong trinh c6 2 nghiem x ^ X j theo dinh ly viet dao ta c6 nhan tu II. MOT S 6 DANG PHUONG TRJNH VO THL/ONG GAP chung se la: x^ - (xj + X2 )x + Xj .Xj 1. Giai phuong tiinh v6 ty bling phuong phap su dung bieu thuc lien hgp: Ta thuong lam nhu sau: Dau hi^u: + Muon lam xua't hien nhan tu chung trong ^ ( ( x ) ta tru di mot lugng ax + b. + Khi ta gap cac bai toan giai phuong trinh dang: yfu^ + '^g(x) + h(x) = 0 Khi do nhan tu chung se la ke't qua sau khi nhan lien hgp ciia Ma khong the dua ve mpt an, hoac khi dua ve mot an thi tao ra nhihig # 0 - ( a x + b) phuong trinh bac cao dan den vi^c phan tich hoac giai true tiep kho khan. + Nham dugc nghi^m cua phuong trinh do: bang thii cong (hoac su dung + De tim a, b ta xet phuong trinh: yfUx) - (ax + b) = 0. De phuong trinh c6 may tinh cam tay) axj + b =
Vi 2: G i a i cac phuong trinh: K h i d o Vsx^-l = V 5 ^ = 2 ; ^ 2 x - l = V 2 ^ = l V i vay ta phan tich p h u o n g t r i n h thanh: a) x/x^ - 1 + x = Vx^ - 2 V5x3-l-2 +^ 2 x - l - l +x - l = 0 b) ^ - 2 ^ - ( x - 4 ) N A r ^ - 3 x + 28 = 0 5x^-5 2x-2 Giai: +x - l =0 ' 75x^-1+2-0 3|(2x-l)^ + ^ / 2 ^ + l a) D i e u kien: x > Ta n h a m d u g c ng hiem x = 3 . N e n p h u o n g t r i n h dugc viet lai n h u sau: 5(x2 + x + l) 2 • «(x-l) +1 = 0 \/x2 - 1 - 2 + x - 3 = 7 x ^ - 2 -5 ^ ' 75x^-1+2 ^(2x-lf +^/2^ +l + x-3= De t h a y : N/X^ - 1 + 2 ^ x ^ - 1 + 4 7x^-2+5 Voi dieu Itien x > 3i t h i f c l l ) + + 1>0 x+3 x^ + 3 x + 9 «(x-3) + 1- =0 V5 V5x3-l+2 ^(2x-l)'+^^^ +l . 7 x 2 - 1 + 2 7 x ^ - 1 +4 7x^-2+5 N e n p h u o n g t r i n h da cho c6 n g h i f m d u y nhat x = 2 "x = 3 b) D i e u kien: x € 2; 4 x+3 Ta n h a m d u g c n g h i e m cua p h u o n g t r i n h la: x = 3 . .7x' - 1 +27x2 -1+4 7x^-2+5 Khi do yf^=^^f3^ = V,^[^^ = ^Jn = l T u d o ta CO l o i giai n h u sau: x+3 x^ + 3 x + 9 Ta d u doan: +1 - < 0 (Bang each thay m p t P h u o n g t r i n h da cho t u o n g d u o n g v o i : 7x2-1+27x2-1+4 7x^-2+5 > A C ^ - 1 +1-N/4^ = 2X2-5X-3 x+3 x2 + 3 x + 9 gia t r i X > 7 2 ta se thay • +1 - 0 = > x = 7 t ^ + l . , < 1; , ^ li; , -< 2 ^xA + 1> -r l !^ 5 i j nen iicii , 1 , .-(2x + l) 7t^ + 1 « t'^ + 3t^ + 6t2 + 4t > 0 . D i e u nay la hien nhien d i i n g . N h a n xet: De d a n h gia p h u o n g t r i n h cuol cung v 6 n g h i e m ta t h u o n g d u n g + Ta xet: cac u o c l u g n g co ban: A + B > A v o i B > 0 tir d o suy ra 0 I > 2 x2 + 3x - 1 > 27x^ - 2 » x^ + 2x^ + 7x2 - 6x + 9 > 0 so A , B thoa m a n 7x^-2+5 B>0
(x^ + x)^ + 6x^ - 6x + 9 > 0 . D i e u nay l u o n d u n g . 4 a+b=8a =— T u d o suy ra p h u a n g t r i n h c6 nghiem d u y nhat: x = 3 De CO dieu nay ta can: < 3 -2a + b = 4 20 b) D i e u k i ^ n : x > 7 . b = De d o n gian ta dat ^ =t>^ => x = t^ + T u o n g t y V l 9 - 3 x - (mx + n ) = 0 nh|n x = 1, x = - 2 la n g h i e m . P h u a n g t r i n h da cho t r o thanh: t2 _ 2t - (t^ - 4)7t^-7 - 3t^ + 28 = 0 3t^ - 1 ^ + 2t - 28 + (t^ - 4)7t^-7 = 0 , fm + n = 4 m = — Tuc la < 0 va t'' > 7 nen t a c o 3Vx + 3 - ( x + 5) (x2-x-2) =0 •^'•^^^ t^+2t + 4 4 t ^ + 7 t + 16 ) + ( t ^ - 4 ) >0. 4 -X' - x + 2 -x-" - x + 2 x'=+x-2 = 0 — 3 3N/X + 3+(X + 5) 3^3^19-3x + ( 1 3 - x ) V i v a y p h u o n g t r i n h c6 nghi?m d u y nhat t = 2 x = 8 . N h a n xet: Vi^c dat \/x = t t r o n g bai toan de g i a m so l u g n g da'u can da g i i i p 1 o - x^ - x - 2 n + 1 =0 d a n gian h i n h thuc bai toan . 3 3>/x + 3 + (x + 5) 3 3Vl9-3x +(13-x) N g o a i ra k h i tao Hen h o p d o (t^ - 4) > 0 nen ta tach n o ra k h o i bieu thuc de 1 De thay v o i - 3 < x < — t h i > 0, — — - — ^>0 cac thao tac t i n h toan dup-c d o n gian h o n . 3 3Vx + 3 + ( x + 5) 3 3Vl9-3x +(13-x) V i d y 3: G i a i cac p h u o t i g t r i n h : 2x-ll N e n —. + 1. a) 4Vx + 3 + V l 9 - 3 x = x ^ + 2 x + 9 b) V3x - 8 - Vx + 1 = 3 3N/XT3+{X + 5) 3\3^fl9^ + {13-x) x^+5x^+4x + 2 ri x=l c) x2+7 d) = Vx +X + 72 P h u o n g t r i n h da cho t u o n g d u o n g v o i x^ + x - 2 = 0 x = -2 '^^x 2(x + l ) x^ + 2 x + 3 Giai: Vay p h u o n g t r i n h c6 2 nghiem la: x = 3, x = 8 . I' • 19 N h a n xet: N e u da nha m dugc hai nghiem ciia p h u o n g t r i n h va d u doan a) D i e u k i ^ n : - 3 < x < — dugc p h u o n g t r i n h chi c6 2 nghiem t h i ta c6 the giai theo m p t each khac 3 n g i n g o n h o n n h u sau: Ta n h a m duoc 2 nghi em la x = l , x = - 2 nen ta p h a n tich de tao ra nhan t u chung la: x^ + x - 2 . De l a m dug-c dieu nay ta th^c hien t h e m b o t nhan t u Xet ham so f(x) = 4Vx + 3 + V l 9 - 3 x - x^ - 2x - 9 tren -3-^ n h u sau: '3 19 + Ta tao ra 4\Jx + 3 - (ax + b) = 0 sao cho p h u o n g t r i n h nay nhan x = 1, x = - 2 la Ta thay x = - 3 ; x = - - k h o n g phai la nghiem, i3 nghiem.
Cty TNHH MTV DWH Khattg Viet 19 -9 Tren -3; ta CO /X + 1 f'(x) = ^ •-2x-2;f'(x) = - -25V3x-8 + 3 x - 4 - 9 ( x + 7 + 5Vx + l ) < 0 ^ -3x ' ^(x + 3 f ^(19-3x) < : : > 3 x - 8 - 5 V 3 x - 8 + —+ — + x + 4 5 V x T T > 0 4 4 nen f(x) = 0 c6 toi da hai nghiem tren '3) V3x - 8 - — + + X + 45v'x + l > 0 . Dieu nay la hien nhien d u n g . x=l M a t khac: f ( - 2 ) = f(l) = 0 nen p h u o n g t r i n h c6 d i i n g hai n g h i f m la x = -2 Vay p h u o n g t r i n h c6 2 nghiem la: x = 3, x = 8 . Chuy: b) D i e u k i ^ n : x > —. N h u n g danh gia de ke't luan A ( x ) < 0 t h u d n g la n h i r n g bat dang thuc khong 3 P h u o n g t r i n h d u g c viet lai n h u sau: 5 V 3 X - 8 - 5 N / 5 ^ = 2X-11 chat nen ta luon dua ve dugc tong cac bieu thuc b i n h p h u o n g . Ta n h a m d u o c 2 nghiem x = 3, x = 8 nen suy ra n h a n t u c h u n g la: N g o a i ra neu t i n h y ta c6 the thay: 5V3x - 8 + 3x - 4 - 9 ( x + 7 + sVx + 1) < 0 x^ - l l x + 24 5\J3x - 8 + 3x - 4 < 9x + 63 + s V s i x + 81 N h u n g dieu nay la hien nhien d u n g Ta phan tich v o i n h a n t u 5\/3x-8 n h u sau: do: 5 V 3 x - 8 < 5 7 8 ] x + 8 1 ; 3 x - 4 < 9 x + 63 v o i m g i x > - ,,,, + Tao ra sVSx - 8 - (ax + b) = 0 sao cho p h u o n g t r i n h nay n h a n x = 3,x = 8 la 3 nghiem. N g o a i ra ta cung c6 the giai bai toan theo each d u n g h a m so 6 cau a) c) Dieu kien: x > 0 j, 3a + b = 5 ja=:3 Tuc la a,b can thoa m a n he: Ta n h a m d u g c x = 1; x = 3 nen bie'n d o i p h u o n g t r i n h n h u sau: 8a + b - 20 ^ b = -4 x2+7 Ta c6: k h i x = 1 = 2,khi x=3 = 2 nen ta t r u 2 vao 2 ve 3m + n = 10 m=l 2(x + ] ) 2(x + l ) + T u o n g t u v o i 5vx + 1 - ( m x + n) = 0 ta t h u dugc: • 8m + n = 15 n=7 Vx^ + 3 - 2 V x x^-4x +3 P h u o n g t r i n h da cho t r o thanh: thi t h u dugc: J x + — - 2 = ^, -2 ^ 5x + 7 Ta c h u n g m i n h : A ( x ) < 0 tuc la: ; = Vx^ + X + 2 « x + 3 - V x ' +2x + 3 =0 x-+2x +3 x^+2x + 3
Tai lifu 6n thi d^i hQC sang t^o vd giai FT, bat l^l, WPT, bal VI - Nguyen i rung r^ien Vang Viet x^ - x - l = 0 • (5x + 7) , , =0 ^ [ ( x + 3) + V x 2 + x + 2 x2+2x + 3 j x+1 + =0 VB-SX^ +(2-x) '(5x + 7) = 0 Xet X +1 + = 0(x + l ) N/S-SX^ +(2-X) + 4 = 0 = 0 (1) •3x^ + ( 2 - X ) (x + 3) + \/x^ + X + 2 x 2 + 2 x + 3^ o ( x + l ) V 8 - 3 x ^ - X 2 + X + 6 = 0 O 2 ( X + 1 ) V 8 - 3 X 2 - 2 X 2 + 2 X + 12 = 0 1 1 Giai (1): •x' + X - V x ^ + x + 2 = 0 . -;2 (x + 3) + \ / x 2 + x + 2 x2+2x + 3 o ( x + l)2 +2(x + l ) V 8 - 3 x 2 + 8 - 3 x ^ + 3 = 0 . X +1 + Vs - 3x + 3=0 D a t t = Vx^ +x + 2 > 0 . P h u o n g t r i n h t r o thanh: P h u o n g t r i n h nay v 6 nghiem. t=2 x=l i + Vs t -t-2 = 0 ^ o x + x - 2 = 0 Vay p h u o n g t r i n h ban dau t u o n g d u o n g v o i x^ - x - 1 = 0 x = t = -l(L) x = -2 15 Ket l u a n : P h u o n g t r i n h c6 3 nghiem: x = — r ; x = l ; x = - 2 b) D i e u kien: 2x3 - 1 5 > 0 x > Vi du 4: Giai cac phuong trinh: Ta vie't lai p h u o n g t r i n h thanh: 2x(x-' - 7) - 3{x^ - 7) - 5 = 4(7 - x^)^J2ix^ - 7 ) a) x^-3x + l = \/8-3x^ Dat t = (x^ - 7) => X = + 7 voi t > - b) 2x^ - 3x3 _ -,4^ ^ |g ^ - 4x^)72x3 - 1 5 2 P h u o n g t r i n h da cho t r o thanh: 2t\/t + 7 - 3t - 5 + 4 t V 2 t - l = 0 Giai: Ta nham d u g c t = 1. N e n phan tich p h u o n g t r i n h n h u sau: a) Dieu k i f n: I x 1< — 3 2 t ^ t T 7 - 4t + 5t - 5 + 4 t 7 2 t ^ T - 4t = 0 D i i n g may t i n h bo t u i ta t h u dugc 2 nghiem la: x^ - - 0 , 6 1 8 , X 2 =1,618 « 2 t f - V r f 7 - 2 ) + 5 ( t - l ) + 4t(V2t-l -1) = 0 Tu d o ta thay x^+X2=l, Xj.X2=-l nen n g h l d e n n h a n t u c h u n g la t-1 2t-2 x-^ - x - 1 »2t + 5 ( t - l ) + 4t =0 U2t-l+lj Xet Vs-Sx^ - ( a x + b) = 0 sao cho p h u o n g t r i n h c6 2 n g h i e m X j =-0,618, X 2 = 1,618 ta t h u duoc: 2t 4t «(t-l) + 5- =0 7(t + 7)2 + 2 ^ / t 7 7 + 4 721^ +1 axj + b = y 8 - 3X] -0,618a + b = 2,61 Ia = - 1 l,618a + b = 0,38 ' ^ ] b =2 2t 4t axj + b = ^ 8 - 3x2 De thay v o i t > — t h i • + 5- >0. 2 7(t + 7)2 + 2 ^ + 4 V2F^+1 P h u o n g t r i n h dug-c vie't lai n h u sau: x'' - 2x - 1 = \/8-3x^ - (2 - x) N h u vay p h u o n g t r i n h c6 nghiem d u y nhat t = 1 x = 2 -4(x^ - x - 1 ) o(x^ - x - l ) ( x +l) = - ^ Vi du 5: Giai cac phuong trinh: 3x^ + ( 2 - x ) a) 2V4x' - X + 1 + 2x = 372x^ - x^ + 79x^ - 4x + 4 4 «>(x^ - x - 1 ) x + 1 + b) V3x + 1 - Vx + 3 - X +1 =0 V8-3x2 +(2-x)^
Tai l.cn 6nthUaihocshngtaovagtdtPl,batl'l,hel^l,bai VI -N^yen imngi^ien Xetham so f(t) = Vt^ + 8 - Vt^ +15 - 3 t + 2 v a i t < - l Giai: a) Phuang trinh duoc viet lai nhu sau: Tac6:f'(t) = -3 + 3t^ < 0 do do f(t) > f ( - l ) = 2 Vl6x2 - 4x + 4 + 2x = 3\/2x^ - + N/QX^ - 4x + 4 ;.Vy " I .Vt^+8 Vt^+15. Suy ra phuong trinh v6 nghiem. %' ' « V l 6x2 - 4x + 4 - Vl 6x2 - 4x + 4 = 3\/2x2 - x^ - 2x > 0 Tom lai: Phuang trinh ban dau c6 2 nghiem la x = 0, x = 1 ,' _ Suy ra dieu kien de phuang trinh c6 nghiem la 3^/2x^ - x'' - 2x > 0 Nhan xet: Trong mpt phuong trinh c6 chiia nhieu dau ta nen dat an phu la mot bieu thuc chua sao cho v i f c bieu dien x theo an do la don o 27(2x2 - x ^ ) > ^ x ^ » 35x3 - 5 4 x 2 < 0 o 35 X < gian nhat. Vi^c la nay se giiip cau true phuang trinh mod d a phuc t^p hon. 54 Xct X = 0 thoa man phuong trinh. b) Dieu ki^n x e -'•'-1 Xet X ?i 0. Ta viet lai phuong trinh nhu sau: sfsx + l - Vx + 3 +1 - x = 0 Trucmg hop 1: x > 0 Chia hai ve phuang trinh cho x ta thu dugc: 2x-2 ;7X4.2=33/1:;.J9-i.4. X X^ . V3x + l + V x + 3 +l-x =0o(2x-2) ^ ' , - ± =0 W 3 x +l+Vx +3 2j x=l De don gian ta dat ^J- - 1 = t - - 1 = t ^ Phuong trinh da cho tro thanh: V3x +1 + Vx + 3 = 2 Vt'^+15+2 = V t ^ + 8 + 3 t o V t ^ + 1 5 - V t ^ + 8 = 3 t - 2 . Xet phuong trinh: >/3x + l + Vx + 3 = 2 . Binh phuang 2 ve ta thu dugc: 2 4x + 4 + 2J(3x + l)(x + 3) = 4 o J(3x + l)(x + 3) = - 2 x o i ' ^ f ° * De phuang trinh c6 nghiem ta can: 3 t - 2 > 0 o t > - . Nham duoc t = 1 nen [x2-10x-3 = 0 O X = 5-2N/7 ' ta viet lai phuong trinh thanh: Vt^ +15 - 4 = - 3 + 3t - 3 Ket luan: Phuang trinh c6 2 nghiem la x = 1, x = 5 - 2\l7 «(t^l) 't-Vl)(t2 +t +l) ( t 3 + l ) ( t 2 + t + l )- 3 =0 Vt^78 + 3 Nhan xet: 15+4 + Ta thay phuang trinh c6 n g h i f m x = l . Neu ta phan tich phuong trinh (t^+l ^2 + t +1 ^2 +t+l thanh V s x + T - 2 + 2 - V x + 3 + 4 - 4 x = 0 thi sau khi lien hgp phuang trinh De y rang:
2. Dat an p h y d y a vao tinh d i n g cap ciia p h u o n g trinh: n=l m = -l Ta t h u a n g gap p h u a n g trinh dang nay 6 cac dang bien the nhu: m - 2n = - 3 n = l 2m + 4n = 2 + a x ^ + b x + c = d^/px^+qx + r (1) t P huong trinh da cho c6 dang: + ax^ + bx + c = d-Jpx"* + qx^ + rx^ + ex + h (2) .j;.,.^, -r,,, iq , -2(x + 2) + 2(x^ - 2x + 4) - 3yJ{\ 2)(x'^ - 2x + 4) = 0 . Chia p h u o n g trinh cho + AVax^ + bx + c + B^ex^ + gx + h = C^/rx^ + px + q (*) •'' x+2 (x + 2) - 2x + 4 > 0 ta thu dugc: - 2 -3 + 2 =0 Thuc chat p h u o n g trinh (*) khi binh p h u o n g 2 ve thi xuat hi^n theo dang (x^ - 2 x + 4) (1) hoac (2). De giai cac p h u o n g trinh (1), (2). Dat t = — — > 0 ta thu duoc p h u o n g trinh: -2t^ - 3t + 2 = 0 P h u o n g p h a p chung la: ^ ( x ^ - 2 x + 4) + Phan tich bieu thuc trong dau V~ thanh tich ciia 2 da thuc P(x),Q(x) t = -2 1 d o t > 0 => t = 1- (^ + 2) 1 2 ^ 4 A, '^S t=- 2 y ( x 2 - 2 x + 4) 2 + Ta bien doi ax^ + bx + c = mP(x) + nQ(x) bang each dong nhat hai ve. ^ ' = — X - 2x + 4 = 4(x + 2). Khi do p h u o n g trinh tro thanh: mP(x) + nQ(x) = d^/P(x)JQ(x) X = 3 + N/I3 Chia hai ve cho bieu thuc Q(x) > 0 ta thu dup'c p h u o n g trinh: x' - 6 x - 4 = 0 o x = 3-Vl3 m —P(x)^ +, n„ = d^ J fPix) — ^ . Dat t = f P W > 0 thi thu dugc p h u o n g trinh: fx>0 02 + S mt^ - dt + n = 0. Binh p h u o n g 2 ve'ciia p h u o n g trinh ta thu dugc: Mpt each tong quat: Voi nriQi p h u o n g trinh c6 dang: x^
Cty TNHH MTV DWH Khang Viet t=- l -4x + l 1 » 1 t = — Giai: Phuong trinh 5t^ + 4t - 1 = 0 t =- 5 x^ +2x + l 25 5 a) Dieu kien x > - ^1 Phuong trinh da cho dugc viet lai nhu sau: 24x^ -102x + 24 = 0 2x^ - 8x2 + 1 0 x - 4 - 3 x ( x - 2 ) V 2 x - l = 0 x =4 o (X - 2)(2x2 - 4x + 2) - 3x(x - 2 ) V 2 x - l = 0 Ket luan: Phuong ^rinh c6 2 nghi^m -x = - , x = 4 c > ( x - 2 ) (2x2 _ 4 x + 2 ) - 3 x V 2 x - l = 0 Nhan xet: Trong lai giai ta da bien doi: x-2 = 0 (x + l)Vx^ - 4 x + l = J(x^ +2x + l)(x^ - 4 x + l ) la vi X +1 >0 (2x2 _4x + 2 ) - 3 x V 2 x - l =0 c) Dieu ki?n: x > -1 Xet phuong trinh: Ta Viet lai phuong trinh thanh: (x - 1 ) 2x^ - 2x - 2 - 3xVx + l = 0 2 x 2 - 4 x + 2 - 3 x V 2 x - l =02x2 - 4 x + 2 - 3 V x 2 ( 2 x - l ) = 0 x=l Jm = 2 Ta gia su: 2x2 - 4x + 2 = ^^^2 ^ ^^2x -1): n = -2 2x2 - 2 X - 2 - 3 X N / X + T = 0 Xet phuong trinh: 2x^ - 2 x - 2 - 3 x V x +1 = 0 2x2 - 3xVx +1 - 2 ( x +1) = 0 . Phuong trinh tro thanh: 2x2 _ 2(2x -1) - 3 ^ x 2 ( 2 x - l ) = 0 . Chia cho x^ > 0 De thay x = -1 khong phai la nghi^m. Ta c6: 2 - 2 . '2x-l' J'—2 x--—1 = 0 . Dat t = '2 x - l / — - — > 0 phuong trinh moi Xet X > -1 ta chia cho x +1 thi thu dugc phuong trinh: X la: -2 (1) TxTT t = -2 2^^ 3 - ^ ^ - 2 = 0 -2t2 - 3 t + 2 = 0 < » 'x + 1 Vx + 1 x = -1 (2) .=1 2 X>0 x=4+2V3 Giai (1): -2C:> o x = 2 + 2^/2 Voi t = i taco: . F ^ = i c : > x 2 - 8 x + 4 = 0 o -4x-4 = 0 x = 4-2>/3 Nhan xet: x
T n i lieu on thi iliii hoc •.mix tao vii giai PT, hat PT, he PT, boT DT- Nguyen TrungKien Cty TNHH MTV DWH Khang Viet -3x-18>0 T o m lai: P h u o n g t r i n h c6 2 n g h i f m la: x = ^ va x = 9 b) D i e u ki^n: x>0 x>6. 5x^ + 4 x > 0 c) D i e u ki?n x > 5 . C h u y e n ve b i n h p h u o n g ta dugc: 2x2 - 5x + 2 = S^^x^ - x - 20J(x + 1 ) P h u o n g t r i n h da cho dvtqc viet lai thanh: Vsx^ + 4x = Vx^ - 3x - 1 8 + 5^/x B i n h p h u o n g 2 v e v a t h u gpn ta dugc: 2x^ - 9x + 9 - 5yjx{x^ - 3 x - 1 8 ) = 0 Gia sir: 2x2 _ 5 x + 2 = m | x 2 _ x - 2 0 J + n ( x + 1 ) N e u ta gia s u 2 x ^ - 9 x + 9 = mx + n(x^ - 3 x - 1 8 ) t h i m , n phai thoa m a n m-2 ' n =2 K h i do ta CO : - m + n = -5 k h o n g ton tai m , n thoa m a n h$. m - 3n = - 9 dieu nay la hoan toan v 6 l y . -20m + n = 2 -18n = 9 N h u n g ta c6 : (x^ - x - 20J(x +1) - (x + 4 ) ( x - 5 ) ( x + 1 ) = (x + 4)(x2 - 4x - 5 De khac phuc van de nay ta c6 chii y sau : x^ - 3x - 1 8 = (x - 6)(x + 3 k h i do m =2 A / X ( X ^ - 3X -18) m =2 = 7x(x - 6)(x + 3) = 7(x^ - 6x)(x + 3) Gia sir: 2x2 - 5x + 2 = a^x^ - 4x - sj + p(x + 4 ) . c,^^y -4m + n = -5: n =3 Bay g i o ta viet lai p h u o n g t r i n h thanh: 2x^ - 9x + 9 - 5yj(x^ - 6 x ) ( x + 3) = 0 - 5 m + 4n = 2 ,m = 2 Ta viet lai p h u o n g t r i n h : 2^x2 - 4x - sj + 3 ( x + 4) = 57(x2 - 4 x - 5 ) ( x + 4) . m =2 Gia sir: 2x2 - 9x + 9 = ^^^^2 _ ^ ^ 3^. - 6 m + n = - 9 n =3 n =3 x2-4x-5 _Jrx2-4x-5] Chia hai v e c h o x + 4 > 0 ta t h u duoc: 2 + 3= 0 N h u vay p h u o n g t r i n h t r o thanh: x+ 4 f x+ 4 J 2{x^ - 6x) + 3(x + 3) - 5^{x^ - 6x)(x + 3) = 0 t=l x2-4x-5^ Dat t = > 0 ta thu d u o c p h u o n g t r i n h : 2t2 - 5t + 3 = 0 x^ - 6 x x^-6x x+4 Chia cho x + 3 > 0 ta t h u dugc: 2 -5 +3=0 2 x+3 x+3 5 + 761 t=l x=• 2 x^ - 6 x T r u o n g h ( ? p l : t = l^^ = 1 o x 2 - 5 x - 9 = 0 Dat t = , > 0 = > 2 t ^ - 5 t + 3 = 0
Giki: l _ 3 ( ^ . 2 i ^ = 0. a) Dieu ki?n: —• Binh phuong 2 ve phuong trinh ta thu dirge: t= - i Dat t = ^ ^ ^ ^ ta CO phuong trinh: 2t^ - 31^ +1 = 0 » 2 + 4x - 1 + 27(x^ +2x)(2x-l) = Sx^ + 4x +1 o x^ +1 - ^(x^ +2x)(2x-l) = 0 Ta t=l m =l m =l Truong hgp 1: * = gia sit: x^ +1 = m(x^ + 2x) + n(2x -1) o < n = - l n =- l 2m + 2n = 0 Jx + 2 1 „ I r x0 = ! \/x + 2= x x^ - x - 2 = 0 «>x = 2 2x-l Dat t = >0=>-t^-t +l =O o t = Ix^ +2x Ke't luan: Phuong trinh c6 2 nghiem: x = 2;x = 2 - 2^3 Ve CO ban den day ta hoan toan tim dugc x. Nhung voi gia tri t nhu vay V i du 4: Giai cat phuong trinh: viec tinh toan se gap kho khan. De khac phuc ta c6 the xu ly theo huong khac nhu sau: a) 2 X ^ - X 2 - 3 X + 1 = N/X^ + X ' * + 1 b) 5 N / X ' ' + 8 X = 4 X 2 + 8 Giai: Ta viet lai: ^(x^ + 2 x ) ( 2 x - l ) = 7(x + 2)(2x^ - x ) liic nay bang each phan tich a) Hinh thuc bai toan de lam cho nguoi giai bo'i roi nhung de y that ky ta thay: nhu tren ta thu dugc phuong trinh: Chia khoa bai toan nam 6 van de phan tich bieu thuc: x^ + x"* +1 1 Ta thay do ve trai la bieu thuc bac 3 nen ta nghi den huong phan rich: 1(2x2 - X) + i ( x + 2) - J(x + 2)(2x2-x) = 0 « ^ ^ ^ ^ - 2 . p ^ ^+1=0 2^ ' 2^ ' ' x+2 V x+2 x^ + x"* + 1 = (x^ + ax + l)(x'' + bx^ + cx +1). Dong nhat hai ve ta thu dugc: a = 1; b = 0; c = - 1 . Nen ta viet lai phuong trinh da cho thanh: Dat t = J— ^ > 0 = > t 2 - 2 t + l = 0 < » t = lci>2x2-x = x + 2x2-x-l = 0 V x+2 2(x^ - X + 1) - (x^ + X + 1) - ^(x^ - X + ]).(x2 + X +1) = 0 1 •m -2 . X^ + X + 1 X^ + X + 1 Ta viet lai phuong trinh thanh: x^ - 3x(x + 2) + lyjix + lf =0 t=l x-" - X + 1 De y rang: Dat t = > 0 ta CO phuong trinh: 2t - 1 - 1 = 0 X^ + X + 1 t = -f(L) Neu ta dat y = Vx + 2 thi phuong trinh tro thanh: x^ - 3xy^ + 2y'^ = 0 . Day la mot phuong trinh d i n g cap bac 3. T u djnh huong tren ta c6 loi giai cho x=0 X - X + 1 , -i 2 ^ bai toan nhu sau: x=- l Giai t = 1 — = 1 x-^ - x^ - 2x = 0 « + Xet truong hop: x = 0 khong thoa man phuong trinh: x^ +X + 1 x=2 1! V + Xet X 0. Ta chia phuong trinh cho x'^ thi thu dugc: Ke't luan: Thu lai ta thay 3 nghiem: x = 0,x = - l ; x = 2 deu thoa man.
Cty TNHH MTV DWH Kltattg Vift ml im on tm a^t H^J sang t4d tagtat m. bat l^l, Vi, uai trr^ Nguyen TnmgKien Giii: X >0 b) D i e u ki?n: + 8x > 0 » a) Dat N/2X + 3 = a , V x + l = b = > a , b > 0 x (a - 2b)(a - b)(a + b - 1 ) = 0 = ( x 2 - 2 x + 4)(x2+2x) N/2X + 3 - Vx + 1 = 0 m +n =4 • ' Ta q u y bai toan ve giai 3 p h u o n g t r i n h co ban la: 2N/2X + 3 -N/X + 1 = 0 Gia s u 4x2 + 8 = ^ ^ ^ 2 _ 2x + 4) + n{x^ + 2x) - 2 m + 2n = 0m = n = 2 N/2X + 3 + V X + 1 - 1 = 0 4m = 8 V o i dieu kien: X > - 1 => a > l , b > 0 - ' P h u o n g t r i n h tro thanh: T r u o n g hop 1: N/2X + 3 - Vx + 1 = 02x + 3 = x + lx = -2(L) 2(x2 - 2x + 4) + 2(x2 + 2x) - 5^{x^ - 2x + 4)(x2 + 2x) = 0 . Chia hai ve cho T r u o n g hop 2: lyJlx + 3 - Vx + 1 = 08x + 12 = x + lx = - y (L) x^+2x x^ + 2 x - 2 x + 4 > 0 ta t h u dug-c: 2- -5 + 2 = 0. -2x + 4 Vx2-2x + 4 T r u o n g hop 3: N/2X + 3 + %/x7T - 1 = 0 . V i N/2X + 3 > 1,N/X + 1 > 0 => VT > O t= 2 Dau bang xay ra k h i va chi k h i x = - I . ^ ^"^^ > 0 ta CO p h u o n g t r i n h : 2 t 2 - 5 t + 2 = 0 « Dat t = x^ - 2 x + 4 ,.1 T o m lai p h u o n g t r i n h c6 n g h i f m d u y nha't x = - 1 2 b) Dieu kien x > - 2 x^ + 2 x Ta thay rang ne'u b i n h p h u o n g true tie'p se dan den p h u o n g t r i n h bac 5 T r u o n g h(?p 1: t = 2 o ^ ' ^" = 4 o Sx^ - lOx +16 = 0 v 6 n g h i ^ m X -2x + 4 De khSc phuc ta se t i m each tach x2 + 4 ra k h o i V2x + 4 tj • -5-V37 T u do ta viet lai p h u o n g t r i n h n h u sau: (x2 + 4)\/2x + 4 + x2 + 4 = 4x2 ^ X=• T r u o n g h g p 2: t = i « - i ^ - t ^ =lo3x2+10x-4 = 0/37 X = • o (x2 + 4)(V2x + 4 + 1) = 2x(V2x + 4 + l)(V2x + 4 - 1 ) 3 Ket luan: P h u o n g t r i n h c6 hai n g h i ^ m la: D o 72x + 4 + 1 > 0 . P h u o n g t r i n h da cho t u o n g d u o n g v o l X =• o x2 + 4 = 2x(V2x + 4 - 1 ) x2 + 2x + 4 - 2xV2x + 4 = 0 o (x - V2x + 4 ) =0 N h a n xet: Ta c6 the phan tich: i X +8x = J x X ^ + 8) = 7x(x + 2)(x2 - 2x + 4) = 7(x2 + 2x)(x2 - 2x + 4) o X = V2x + 4 o x>0 o x = l + V5 x^ - 2 x - 4 = 0 Chu y r5ng: T r o n g m p t so p h u o n g t r i n h : Ta can dua vao tinh ddng cap cua timg nhom so'hang detit do phan tich tao thanh nhan tu chung. Ket luan: P h u o n g t r i n h c6 n g h i ^ m d u y nha't x = 1 + N/S c) D i e u ki?n: x > 2 V i dv 5: G i a i cac p h u o n g t r i n h : fm = l a) (x + 2)(V2x + 3 - 2Vx + 1 ) + ^2x2 + 5x + 3 - 1 = 0 Gia sir x2 - 6 x + l l = m(x2 - x + l ) + n ( x - 2 ) - m + n = - 6 m = 1, n = - 5 b) (x^ + 4 ) V 2 x + 4 =3x2 + 6 x - 4 m-2n = l l
Tai lieu oil thi dai hoc sang tao va giai PT, bat PT, hf PT, boT DT - NguyettTrung Kien + De giai cac phuong trinh dang nay ta thuong lam theo each: p= l - Dat 7f(x) = t => t^ = f(x) x2 - 4 x + 7 = p ( x 2 - x + l ) + q ( x - 2 ) : - p + q = -4 p = 1, q = -3 p-2q = 7 - Ta t^o ra phuong trinh: mt^+g(x)t + h(x) = 0 Phuong trinh da cho tro thanh: Ta CO A = [ g ( x ) ] ^ - 4 m . h ( x ) = fj(m)x^+gj(m)x + h j ( m ) . De A eo dang (x^ - x + l ) - 5 ( x - 2 ) V x ^ - x + 1-2 (x^ - x + l ) - 3 ( x - 2 ) N/X^ = 0 A(x)]^ thi dieu ki§n can va dii la A^j, = [ g i ( m ) ] ^ -4£j(m).gj(m) = 01=> m Chia phuong trinh cho ^(x^ - x +1)^ ta thu duQc: Ta xet cac vi du sau: Vi 1: Giai cac phuong trinh: x-2 x-2 x-2 + 6, ^ I 1-5.- • -21— =0 a) x ^ + l - ( x + l)\/x^-2x + 3 = 0 b) 2V2x + 4 + 4 ^ 2 ^ = 79x^+16 x^-x + l Vx^-x + l V ,x - x + 1 Giai: D3t t = ^ >0 a) Dat t = V x ^ - 2 x + 3 > 0 = > t ^ - x 2 - 2 x + 3 Vx^-x + l Phuong trinh da cho tro thanh: x^ +1 - (x + l)t = 0 '' t= l Ta se tgo ra phuong trinh: mt^ - (x + l)t + x^ +1 - m(x^ - 2x + 3) = 0 Ta thu duoc phuong trinh: 6t^ - 5t^ - 2t +1 = 0 t = l (Ta da them vao mt^ nen phai bot di mpt lupng mt^ = m(x^ - 2x + 3)) 3 1 Phuong trinh dupe vie't lai nhu sau: t= ~(L) mt^ - (x + l ) t + (1 - m)x^ + 2mx +1 - 3m = 0 + Neu t = l « a = b o x 2 - 2 x + 3 = 0(VN) A = (x +1)^ - 4m (1 - m)x^ + 2mx +1 - 3m + Neu t = -x^-10X + 19 = 0 < » X = 5±N/6 3 = (4m^ - 4m + l)x2 + (2 - Sm^)x + 12m2 - 4m +1 Ketluan: X = 5±N/6 Ta mong muon 2. Giai phuong trinh v6 ty bang phuong phap dat an phy khong hoan toan. A = (Ax + B)2 o A ^ = (1 -4m2)2 - { U m ^ - 4 m + l){4m^ - 4 m + l ) = 0c:>m = l + Dat an phu khong hoan toan la phuong phap chpn mpt so hang trong Phuong trinh moi dupe tao ra la: t^ - (x + l)t + 2x - 2 = 0 phuong trinh de dat lam an sau do ta quy phuong trinh ban dau ve dang Ta CO A = x^ - 6x + 9 = (x - 3)^ mpt phuong trinh bac 2: mt^ +g(x)t + h(x) = 0 (phuong trinh nay van con an x ) t_x + l-(x-3)_, Tu do ta c6: 2 + Van de ciia bai toan la phai chpn gia trj m bang bao nhieu de phuong trinh t . i i ± i ± i ^ = x-l bac 2 theo an t c6 gia trj A chan A = A(x) nhu the viec tinh t theo x se dupe de dang. + Truonghppl: t = l o > / x ^ - 2 x + 3 - 2x^-2x-l = 0 o x = l ± r y 2 + Thong thuong khi gap cac phuong trinh dang: ^ , fx>i ax^ + bx + c + (dx + e)^jp\^ + qx + r = 0 hay + Truong hpp 2: t = x - 1 o Vx -2x + 3 = x - l -2x + 3 = x^-2x + l ax^ + bx + c + (dx + e)^px + q = 0 thi phuong phap dat an phu khong hoan Phuong trinh v6 nghiem. toan to ra rat hi^u qua: Tom lai: Phuong trinh c6 2 nghiem la: x = 1 ± %/2
b) D i e u k i # n : - 2 < x < 2 Dat t = Vl - x2 ta tao ra p h u o n g t r i n h : B i n h p h u o n g 2 ve p h u o n g t r i n h va t h u gpn ta dugc: mt2 + (6x + 18)t + (8 + m)x2 - 6x - 18 - m = 0 _ 1 6 7 8 - 2 x 2 + 8x - 32 = 0 . Co A' = (3x + 9)^ - m (8 + m)x'^ - 6 x - 1 8 - m v£ 5 ! v Dat t = yjs-lx^ ta tao ra p h u o n g t r i n h la: = ( 9 - 8 m - m 2 ) x 2 + ( 5 4 + 6m)x + m 2 + 18m + 81 '' ^ mt^ - 1 6 t - m ( 8 - 2 x 2 ) + 9x2 + 8 x - 3 2 = 0 Ta m o n g m u o n o mt2 - 16t + (9 + 2m)x2 + 8x - 8 m - 32 = 0 A = ( A x + B)2 A ' ^ = ( 3 m + 27)2 - ( 9 - 8 m - m 2 ) ( m 2 + 1 8 m + 81) = 0 A' = 64 - m (9 + 2m)x2 +8x-8m-32 T u do tinh dugc m = - 8 8m2 + 32m + 64 = ( - 2 m 2 - 9m)x2 + 8mx + 8r P h u o n g t r i n h da cho t r o thanh: -8t2 + (6x + 18)t - 6x - 1 0 = 0 Ta m o n g m u o n A' = ( A x + B)2 A = 0 phai c6 n g h i e m kep . Tuc la: Ta CO A' = (3x + 9f - 8(6x + 10) = (3x + 1)2 A.;^ =16m2 - ( - 2 m 2 - 9 m ) ( 8 m 2 + 32m + 64) = 0 o m = - 4 ^_ - 3 x - 9 - ( 3 x + l ) _ 3x + 5 -8 4 T u d o suy ra p h u o n g t r i n h m o i la: -4t2 - 16t + x2 + 8x = 0 Suv ra t - - 3 x - 9 + (3x + l ) _ ^ 8 - ( 2 x + 8) _ x t= -4 ~2 T i n h dugc: A' = 4x2 + + ^4 = (2x + 8)^ 8 + (2x + 8) x , T r u o n g h g p 1: t = 1 Vl - x2 = 1 x = 0 thoa m a n d i e u k i ^ n t= -= 4 -4 2 T r u o n g h o p 2: t = o 4 ^ 1 - x 2 = 3x + 5 « 16(1 - x2) = 9x2 + ^ 25 X / 9~ X x>0 4V2 T r u o n g h o p 1: t = — v 8 - 2 x ="^" 4 ( 8 - 2 x 2 ) = x 16(1 - x 2 ) = 9x2 + 3 0 x + 2 5 0 2 5 x 2 + 30x + 9 = 0 o x = - - + T r u o n g h o p 2: T h u lai ta thay: ^ = ~ ~ thoa m a n p h u o n g t r i n h : x
De y r3ng: 2x + 3 = 4(x + 2) - ( 2 x + 5) = (2Vx + 2 - V 2 X + 5)(2N/X + 2 + 7 2 x + 5J D | t t = 72x^ - 3x + 1 ta tao ra p h u o n g t r i n h : N e n ta c6: m t ^ - 8xt + (10 - 2m)x^ + (3m - 9)x + 3 - m = 0 (2VX + 2 - V2x + 5 ) ( 2 V x + 2 + V2X + 5 ) = (2VX + 2 - V2X + 5 ) Ta CO A = 1 6 x ^ - m (10 - 2m)x^ + (3m - 9)x + 3 - m 2Vx + 2 - V2x + 5 = 0 • T = 16x2 _ ^ ( 1 0 - 2 m ) x 2 + ( 3 m - 9 ) x + 3 - m 2Vx + 2 + V2x + 5 = 1 3 = ( 2 m 2 - 1 0 m + 16)x2 + ( 9 m - 3 m 2 ) x + m ^ - 3 m 2Vx + 2 - V 2 x + 5 = 0 X= — 2 2Vx + 2 + V2x + 5 = 1 Ta can : = ( 9 m - 3m^f - 4(2m2 - 10m + 16){m^ - 3 m ) = 0 => m = 3 x = -2 T n r o n g hgrp 2 : t = x - 1 2\fx + 2 - V2x + 5 = x - 1 P h u o n g t r i n h da cho t r o thanh: 3t^ - 8xt + 4x2 = 0 (2Vx + 2 - V2x + 5)(2Vx + 2 + V2x + 5) = (x - l){2yfx + 2 + V2x + 5 ) t = 2x 0 x>0 t = - x c t . V 2 x 2 - 3 x + l = - x < = > - 9 ( 2 x 2 - 3 x + l ) = 4x2 ^ ^ ^ = 2 V ^ + V2x + 5 < = > 2 V ^ + V2x + 5 - ^ ^ = 0 14x2 - 27x + 9 = 0 x-1 x-1 3 3 3 N h a n thay tren m o i khoang ( - 2 ; ! ) va (l;+oo) x =— o 2 lien tyc va c6 3 h a m so f ( x ) = 2\/x + 2 + -Jlx + S _ x - 1 ^ X= — 7 1 1 f(x) = . > 0 nen h a m so' d o n g bien. Suy ra tren m o i s/x + 2 • +yJ2x=+=5= + - {x-lf — T r u o n g h o p 2: t = 2x < » ^j2x^ -3x + l =2xi'^^^ x = - [-3x + l = 0 3 khoang d o p h u o n g t r i n h c6 nhieu nhat m o t n g h i ? m . 3 3 1 Xet tren khoang ( - 2 ; ! ) ta c6 f ( - 2 ) > 0 nen p h u o n g t r i n h v 6 n g h i f m Ke't l u a n : P h u o n g t r i n h c6 3 n g h i f m : ' ' ^ ^ ' ' ^ ^ y ' ' ^ " ^ Xet tren k h o a n g (l;+oo) c6 f(2) = 0 n e n p h u o n g t r i n h c6 n g h i ^ m d u y nhat x =2 b) D i e u k i e n : x > 1 . 3 T o m lai p h u o n g t r i n h da cho c6 d u n g 3 n g h i f m : x = 2,x = - 2 Dat t = 7 x ^ + 3 > 0 x^ = t2 - 3 . D o h? so cua x^ t r o n g p h u o n g t r i n h la: 1 Vi 3: G i a i cac p h u o n g trinh: P h u o n g t r i n h da cho t r o thanh: t2 - (5x - l ) t + 6x2 - 2x == 0 a) lOx^ - 9 x - 8 x V 2 x 2 - 3 x + 1 + 3 = 0 A = (5x-1)2 -4(6x2 -2x) = x 2 - 2 x + l = ( x - l ) 2 . b) x^ + 6 x ^ - 2 x + 3 - ( 5 x - l ) \ / x ^ + 3 = 0 ^_(5x-l)-(x-l)_ 2x Suy ra: Giai: x>l 2 a) D i e u k i f n: x . l 2
b) Ta viet 1 ^ phuang trinh thanh: 3 - (2x2 _^ + x + x'* = 0 x=l x>0 Ta coi day la phuang trinh bac 2 cua Vs ta c6: 3 + 721 Tmong hgp 1: + 3 = 2x • X=• x ^ - 4 x ^ + 3 = o' 2 A = ( 2 x 2 + 1 ) 2 - 4 ( x + x^) = 4 x 2 - 4 x + l = (2x + l)2 . \ « 3-^/2T x=- (L) > / 3 = - ( 2 x 2 + l + 2 x - l ) = x2 + x x2 + X - Vs = 0 Tir do suy ra x= l x2 - x + l - V s = 0 V3=-i(2x2+l-2x +l)= x2-x + l Truong hg-p 2: 7 x ^ + 3 = 3x - 1 X = 4 + 2N/3 x^ -9x^ +6x + 2 = 0 Giai 2 phuang trinh tr§n ta thu dugc cac nghi?m ciia phuong trinh da cho X = 4-2N/3(L) - l ± V l + 4V3 . „ -1±V4N/3-3 3+ >^ la: x = ^ hoac x = ^ Tom lai phuong trinh c6 3 nghi^m: x = 1, x = ,x = 4 + 3V2 2 2 V i dv 4: Giai cac phuong trinh: c) Dieu ki?n x > - 4 Ta viet lai phuang trinh thanh: x + 4 + (4x2 ^^_2|7x + 4 + 8x2 + 2 x - 8 = 0. a) V s - x =x^ - 5 b) X ' ' - 2 N / 3 X 2 + X + 3->/3=0 Coi day la phuang trinh bac 2 an Vx + 4 thi c) 8X2+3X + ( 4 X 2 + X - 2 ) N / X + 4 = 4 A = (4x2+x-2) - 4 ( 8 x 2 + 2 x - 8 j = 4x2 _^ ^ -4- Gidi: Vx + 4 = -2x Tu do suy ra x5 Giai 2 truong hgp ta thu dugc cac nghi^m cua phuang trinh la: Binh phuong 2 ve ta thu dug-c: 5^ - (2x^ +1).5 + x + x"* = 0 1-V65 Ta coi day la phuong trinh bac 2 cua 5 ta c6: x=- 8 A = (2x2 +1)2 - 4(x + X * ) = 4x2 - 4x.+1 = (2x +1)2 -3 + V57 x=- 8 5 = i ( 2 x 2 + l + 2 x - l ) = x2 + x Vi d\ 5: Giai cac phuong trinh: T u do suy ra 5 = -i(2x2+l-2x +l)= x2-x + l a) 3(V2x2 + 1 -1) = x ( l + 3x + 8V2x2 +1) b) Vx2 +3x + 6 + 7 2 x 2 _ i =3x + i -1-N/2T X =• Giii: Truonghgp 1: x 2 + x - 5 = 0 ~*"2 a) Ta viet lai phuang trinh thanh: 3x2 + x + 3 + (8x - 3 ) ^ 2 x 2 + 1 = 0 . ' x =- 2 D|t t = V 2 x 2 + l > 0 suyra t^=2x^+\. Ta tao ra phuang trinh: mt2 + (8x - 3)t + (3 - 2m)x2 + x + 3 - m = 0 . X =- Truong hgp 2: x - x - 4 = 0o 2 Ta CO A = ( 8 x - 3 ) 2 - 4 m ( 3 - 2 m ) x 2 + x + 3 - m X =• = (8m2 - 12m + 64)x2 - (48 + 4m)x + 4m2 - 12m + 9. Doi chieu voi dieu ki^n ta c6 4 nghi^m deu thoa man phuong trinh.
Ta can A' = (24 + 2mf - (8m^ - 12m + 64)(4m2 - 1 2 m + 9 ) = 0 = > m = 3 . E)6'i chie'u v o l dieu k i ^ n ban dau ta tha'y chi c6 x = -^''"^TlS ^j^^^ man P h u o n g t r i n h t r o thanh: 3t^ + (8x - 3)t - 3x^ + x = 0 . Ta c6: A = (8x - 3)^ - \2.{-3x^ + x) = lOOx^ - 60x + 9 = (lOx - 3)^ . 1 .* dieu kien . ^^3-8x-(10x-3)_3^^^ 6 2x-l x.l X =• 1 Tir d o t i n h dvtqc : T r u o n g h o p 2: ^2x2 - 1 = 2 o ^_ 3 - 8 x + ( 1 0 x - 3 ) _ X 4x2 + 4 x - 5 = 0 X = • 6 ~ 3 1 T r u o n g hgip 1: V2x^ + 1 = - 3 x + 1 x < - X = 0 Do'i chieu v o i dieu k i ^ n ban dau ta thay chi c6 x = ^ la thoa m a n dieu 3 9x2 _ 6 x = 0 ki#n . x PHLTONG P H A P HAM S6 Dau hi?u: Ta viet lai p h u o n g t r i n h thanh: N / X 2 + 3 X + 6=3X + 1-\/2X2-1. + Bai toan p h u o n g t r i n h giai bang p h u o n g phap h a m so' t h u o n g c6 dac d i e m Binh p h u o n g 2 ve'va thu gon ta dugc p h u o n g t r i n h m o i : la l u o n d u a ve d u g c dang: f u(x) = f v(x) trong d o h a m so' dac t r u n g lOx^ + 3 x - 6 - 2 ( 3 x +1)72x2 -1=0 t h u o n g la ham d o n d i e u tang hoac d o n d i ^ u g i a m tren m i e n xac d i n h D Dat t = \/2x2 - 1 > 0 suy ra =2x2-1. + Dac diem noi bat nhat ta c6 the de phat h i f n la: Trong p h u o n g t r i n h c6 nhieu Ta tao ra p h u o n g t r i n h : mt2 - 2(3x + l ) t + (10 - 2m)x2 + 3x - 6 + m = 0 . Ta c6 bieu thuc chua can, hoac da thuc bac cao ma ta k hong the q u y ve m p t an. A' = (3x + 1)2 - m ( 1 0 - 2 m ) x 2 + 3 x - 6 + m Ta t h u o n g giai cac p h u o n g t r i n h dang nay theo each: Cach 1 : = (2m2 - 1 0 m + 9 ) x 2 + (6 - 3m)x - m2 + 6 m + 1 . + D u a p h u o n g t r i n h ve dang f(x) = 0 v o i x 6 D T a c a n A ^ = ( 6 - 3 m ) 2 - 4 ( 2 m 2 - 1 0 m + 9)(-m2 + 6 m + l ) = 0 = > m = 4 . + Xet ham so' y = f(x) tren D . P h u o n g t r i n h t r o thanh: 4t2 - 2(3x + l ) t + 2x2 + 3x - 2 = 0 C h u n g m i n h f'(x) > OVx € D hoac f "(x) < OVx € D Ta c6: A' = (3x + 1)2 - 4.(2x2 + 3x - 2) = x^ 6x + 9 = (x - 3)2 . + N h a m m o t n g h i f m x = Xg . D u a vao t i n h chat ciia h a m so' d o n di?u ta suy ra 3x + l - ( x - 3 ) x+ 2 n g h i e m d o la d u y nhat. t=- Tir do t i n h dugc: 4 2 * T a xet cac vi dvi sau: ^_3x + l + ( x - 3 ) ^ 2 x - l V i dy 1 : G i a i cac p h u c m g t r i n h sau: a) V x - l + x 2 - 7 + ^ x + 6 = 0 2 + 2V15 x>-2 X =• 7 T r u o n g h g p 1: V2x2 - 1 =