Rèn luyện kỹ năng sáng tạo và giải phương trình, hệ phương trình, bất phương trình: Phần 2

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Nối tiếp nội dung phần 1 cuốn Sáng tạo và giải phương trình, hệ phương trình, bất phương trình, phần 1 cung cấp cho người đọc các kiến thức: Phương trình, bất phương trình chứa căn thức; hệ phương trình, hệ bất phương trình; các bài toán phương trình, hệ phương trình, bất phương trình trong đề thi đại học. Mời các bạn cùng tham khảo.

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Nội dung Text: Rèn luyện kỹ năng sáng tạo và giải phương trình, hệ phương trình, bất phương trình: Phần 2

p a l t o a n 2 . Giai cac phuong trinh sau , j,.. ^ a) ^^2^^ + v ' ' ^ ^ = v ^ S ^ I ; 6) sJ^FTT + v ^ ^ ^ = v ^ . Dieu kien x € R. Lap phUdng hai ve phUdng t r i n h da cho t a ditdc Chtfdng 3 ; ; / 2 x - l + x - l + 3 v / 2 r ^ v / ^ ^ ( v ^ 2 7 ^ + S / ^ ^ = 3x + 1 ^^27^. (i) G i a i . Tap xac d i n h R. Ta giai (1) bang each xet dau ham so lien tuc tren .c G i a i . Lap phudng hai ve phildng t r i n h (1) t a dudc ok 15.T - 1 + 13.r + 1 + 3 iJ/(15.T - l)(13x + 1) {^Ihx - 1 + V'13.T + 1) = 64.T. bo M a ^ 1 5 x - 1 + s/13x + 1 = 4^/5 nen Taco ce fa 15x - l + 13a; + l + 3v/(15x - l ) ( 1 3 x + l ) . 4 ^ = 64x (2) w. ^ 1 2 ^ x ( 1 5 x - l ) ( 1 3 x + l ) = 36x ^ ^ x ( 1 5 x - l ) ( 1 3 x + 1) = 3x
+ 2f = 2x + 15. K e t hdp vdi (1) ta c6 he T h a y x = 0, x = - ^ vao (2) thay 3: = - ^ la nghiem d u y nhat ciia ( 2 ) . / (4i/ + 2)2 = 2 x + 1 5 (2) Ham so f{x) lien tuc va phifdng t r i n h 2 1 (4x + 2)2 = 2 t / + 15. ( 3 ) f{x) = 0 CO nghiem d u y nhat tren R. 1 i rQfl thco t i n i g vc cua (2) va (3) t a dUOc Vay /(.T) CO khong qua m o t Ian doi dau ./(-T) 0, / ( - I ) = ^ s/^ + + - - s/=3 < 0. V?ly / ( x ) > , K h i X = y, thay vao (3) dudc (4x + 2f = 2x + 15 ^ / 0 T a p nghiem cua (1) la + 0 0 ^ . 4' IGx^ + 14x - 11 = 0. 01 Giai phUdng t r i n h nay dUdc hai nghiem la x = - va x = . S o sanh v d i oc 1 ,^ dieu kien cua x va y chi c6 nghiem x = - thoa m a n . iH 3.2 Phu-dng trinh [ax + 6)" = p\/a'x + b' + qx + r. Da • K h i 1 + 8(x + y + 1) = 0 y = - X - - , thay vao (3) t a dUdc hi vdi .7: la an so ; p, q, r, a, h, a', b' la cac hang so ; paa' ^ 0 ; n e {2,3}. 8 nT (4x + 2)2 = - 2 x - ^ + 15 0. Q _l_ ^222 Ta ki^n ciia x va y c h i c6 nghiem x = — thoa man. Vay phirdng t r i n h - D a t s/a'x + b' = -{ay + b) neu pa' < 0. s/ lb • B a i toan dan den giai he phurtng t r i n h hai an (thUrJng l a he doi xi'mg loai ... , , . , 1 . - 9 + v/22T up 2) doi v d i x va y. De giai he nay, t a t r i i hai phudng t r i n h (hoac cong hai da cho CO hai nghiem l a x = - va x = . ro phirdng t r i n h ) va do y don t i n h ddn diou ciia ham so. 2 16 Li/u y. Co t h g tiep can bai toan nay thong qua cong cu dao h a m n h u sau : /g • T h u a t dat §,n p h u n h u tren goi la thuat dat "an phu doi xutig". Xet ham so / ( x ) = 32x2 ^ 2>2x - 20. K h i do / ' ( x ) = 64x + 32 = 32(2x + 1). om C h u y 1. Phicanfj trmh dang (ux + ^ ) " = p \/a'x + b'+ qx + r rd tM giai duclr. V%y, dat V2x + 15 = a(2y + 1), t a c6 h? .c phudng phdp nay thiidng thoa man dieu kien : 2x + 15 = a2(2y + 1 ) 2 ^ / + = / ^ ^ ^+ 28 ok • Neu b = 0 ihir = b'. a(2y+l) = 8(2x+l)2-28 ^ 1 2y + 1 = - ( 2 x + 1)2 - — . ( a' -q b'-r , , , , bo a a I p = = —;— (neu pa' > 0) 0, ce • Neu b^O thi b I a' - q b' - r , . , „^ Ttt he nay, de thay rang c i n chon a sao cho a2 = — a b B&i toan 6. Giai phudng trinh 4x2 ^. ^'3^. + j -f- 5 = I 3 x . w. ww 3.2.2 M o t s6 bai toan r e n luyen v a nang cao. GiSi. Dieii kien x > -~. phirdng t r i n h da cho viet lai B a i toan 5 . Giai phuang trinh \/2x + 15 = 32x2 ^ 323. _ 20. o (2x - 3)2 = - V / S ^ M H : + X + 4. (1) G i a i . Dieu kien 2x + 15 > 0 - 7 , 5 . Phifdng t r i n h d a cho viet lai ^ ^ t v'3x + 1 = - ( 2 y - 3 ) , dieu kien - ( 2 y - 3) > 0 y < 1. 5. K h i do v'2x + 15 = 2(4x + 2)2 - 2 8 . (1) ^ + 1 = (2y - 3)2. K e t hdp vdi (1) t a c6 he Dat s/2x + 15 = 42/ + 2, v(3i dieu kien 4j/ + 2 > 0 - 0 , 5. K h i do ( 2 x - 3 ) 2 = 2y + x + 1 (2) • ^ ' ( 2 y - 3 ) 2 = 3x + l (3) 160 161
TT 7n = (JUS 1 •'' — ^ 1 ^ = cos la t a t ca cac nghiem ciia phUdng t r i n h Tiu; theo ve cac p h u d i i g t i i n h (2) va (3) t a dirdc VW H V a jo) va cung la t a t ca cac nghiem cua phUdng t r i n h da oho. « 2(2x + 2y- 6){x - y) = 2y -2x^{x- y){2x + 2y - 5) = 0. T ufu y- Phep dat \/6x + 1 = 2?/ da ditdc noi d phan phUdng phap giai. Tuy • Tritdng hdp y = x. T h a y vao (2) t a durtc ••. - - hien ncu qucn phudng phap t h i t a van t i m ra dirdc phcp dat nay di.ra tren guy luan t u nhien n h u sau : G i a sijf s/6x + 1 = ay + b. K h i do 4a;2 - 12a; + 9 = 3x + 1
i T a CO (4) ^ y = x. T h a y vao (2) t a dUdc ,fiay diicfc gidi bang each diCa ve he "gan" doi xiing loai hai {nghia la khi tril, cQTig hai phuang trinh, cua he ta c6 thUa so [x - y)). Vay ta xet he •> . • (2x -3f = 3x -5 ^ 8x^ - 36x^ + 54x - 27 = 3x - 5 X = 2 . r (2y-5)3 = x - 2 (i) (x - 2)(8x2 - 20x + 11) = 0 ^ 5±V3 • \x - 5)3 = - x + 2y - 2. (ii) ,.. , ^xO'S • ,> 4 rnn (Mttfi jVew CO phep dat 2y - 5 = \/x - 2, thi sau khi thay vao phUdng trinh [ii) dupe 3B^ 'Do 3"^ = > 0 nen (5) khong the xay ra. gj;3 - 60x2 igQ^ _ = - X + ^ x ~2 + 5 - 2 . Ta c6 bai todn A'^ + AB+ {A+--- + sau. 5 i / g a i t o a n 9. Gidi phuang trinh ^x-2 = Sx^ - eOx^ + 151x - 128. 01 PhiTOng t r i n h c6 ba nghiem x = 2, x = —-—. Lvfu y . T h e m m o t phuang phap nfra dc t u n ra phcp dat 2j/ - 3 = v^3x oc Giai. do la : Ta dat ay + b — \/3x - 5, vdi a, b se t i m sau, sao cho cung vdi p h u d n g C a c h 1. PhUdng t r i n h da cho dUdc viet lai ^ x - 2 = ( 2 x - 5)^ + x - 3. (1) iH tririh da cho t a t h u diTdc m o t he doi x i i n g loai I I , hoac la m o t he gan doi Dat 2?y - 5 = v^x - 2. K e t hdp vdi (1), t a c6 he Da xiing loai I I (trijf hoac cong hai phudng t r i n h t a t h u dvtdc x = y ) . T a c6 hi (2y-5)3 = x - 2 (2) (2x - 5)3 = - X + 2y - 2 (3) nT uO Lay (3) trir (2) theo ve t a dUdc - ie Vay t a chon avkb sao cho | 2 ~ ^ ^ { a = 2 ^ T i t do phcp doi bicii 2 (x - y) [(2x - 5)2 + (2x - 5) (2y - 5) + (2y - 5)2] = 2(y - iL x) [ -3-6 = 0 Ta la 2y - 3 = v^3x - 5. C i i n g c6 the tiep can phep d o i bien 2y - 3 = v^Sx - 5 rx-y = 0 (4) s/ [ (2x - 5)2 + (2x - 5) (2y - 5) + (2y - 5)2 + 1 = 0. (5) thong qua cong cu dao h a m nhit sau : X c t ham so up / ( x ) = 8x^ - 36x2 ^ _ • Ta CO (4) y = X . T h a y vao (2), t a dudc ro /g K h i do om (2x - 5)3 = X - 2 8x3 _ gQ^2 ^ j 4 9 ^ - 123 = 0 " /'(x) = 24x2 - 72x /"(x) = 48x - 72 = 24 (2x - 3 ) . (x - 3)(8x2 - 36x + 41) = 0 X = 3. .c 11? si / Vay dat ^ 3 x - 5 = a ( 2 y - 3). T a t h u dUdc he ( -I 302 ok •DoA^ + AB + B'^ = + > 0 nen (5) khong the xay ra. bo 3x - 5 = a3(2y - 3)^ Phudng t r i n h c6 nghiem duy nhat x = 3. / 3x - 5 = a 3 ( 2 y - 3)3 ^ f ce \y - 3) = (2x - 3)3 - X + 2 ^+ \2 ^1 2y-3 = i (2x - 3)^ - X + 2 ^ 0 phuang trinh cd nghiem duy nhat x = 3 nen ta nghi den phuang phdp svt fa a dung tinh dOn dieu cua ham so nhu sau : C a c h 2. Tap xac d i n h E . D a t y = s / J ^ . w. T a c6 h f Tvc he nay, de thay r^ng can chon a sao cho = — =^ a = ±1. ww / 8x3 - 60x2 + i 5 1 x - 128 = y \; = y3 + 2 3.2.3 Phifdng phdp sang tac bai toan mdi. *^Oiig ve theo ve hai phitdng t r i n h cua he, t a ditdc V i d u 1. Ta se sang tdc phicang trinh c6 it nhat mot nghiem theo y muSn- 8x3 _ gQ^2 ^ j 5 2 x - 128 = y3 + y + 2 ' ' Xet X = 3. Khi do 2x - b = 1 => (2x - 5)^ = 1 ' ^ ^ ^ x - 2. Ta mong muon CO mgt phuang trinh chtia {ax + b)^ va ch-da \/cx + d, hon nUa phuong trinh
Xet ham s6 f{t) = e + t. Vi f'{t) =3f + l>Q,Vt&R nen ham / dong biln Jjiy p h U d n g t r i n h (b) trxX p h U d n g t r i n h ( a ) t h e o v e t a d U d c tren E. Do do (*) viet lai f{2x - 5) = f{y) ^ 2 x - 5 = y. Bdi vay {2x - 5) = ^yr^ - 2 . T a t h u d U d c h f (3) (2) CO khong qua mot nghiem. Xet .c r x + 6 = (t/ + 2)2 ^ f x + 6 = (y + 2)2 ok 2=1(^^-1-] ^(a3)2_4^3_i^^3^2±v^. \ + 2 = x^ + 4x \ + 6 = (x + 2)^ 2V bo (4) Lay hai p h U d n g t r i n h c i i a h e (4) trir n h a u , t a d U d c ce Do do, neu dat a = v / F + T ! thi 2 = ^ - Ta c6 fa x-y={y-x){y + x + 4)-!^ y =X w. 1 f 1^ + 4 1 • Khi y = X, t a d U d c x + y + 5 = 0. ww = 3 a- - J .2 V 2 V ^ = x+ 2^{ ^2^32_2^0 ^ x = 3 + VT7 2 Vay X= i - =i (v/2 + v / 2 - \/5) l a n g h i e m duy nhat ciia •^Khi y = -5 - x, ta dUdc p h U d n g t r i n h (2) v a c u n g l a n g h i e m duy n h a t c i i a p h U d n g t r i n h da c h o . C a c h 2. Viet lai (2x)3 = ^ - 6 x + 4 - 8x + 4. Dat 2y = ^^4^^. Ta c6 h e - 5 - \/l3 i ^ ^ ^ = - ^ - 3 ^ { x2-+5x + 3 =0 = 82/'^ = 4 - 6x ^ 8'(/3 - 6 x + 4 (a) Ph uong trinh c6 hai nghiem x = -3 + v ^ -5-v/l3 X = 8x3 + 8 x - 4 = 2y \ 2y + 4 - 8x. (6) 2 166 167
Bai toan 12. Giai phUOng trinh 2x^ - 6 x - 1 = y/4x + 5. •yay vay neu ta chon n = 2 thi se thu dUdc cac phu'dng trinh quy ve bac hai, do Dap so. .r = 2 + v ^ , .T = 1 - \/2. (j5 vdi n = 2, de sang tac bai toan mdi thi viec chon / ( x ) , b{x), a{x) la d§ dang, khong can can nhac nhieu. Tuy nhien vdi n = 3 hoac Idn hdn, de sang Bai toan 13. Giai phuang trinh - 4x - 3 = sjx + 5. ' '' * tac bai toan mdi thi viec chon / ( x ) , b{x), a{x) can phai ditdc hta chon, can f i\ nhac ki lirdng sao cho sau khi dat an phu, dua ve he doi xiing loai I I , dua Dap so. Tap nghiem S = ^(5 + v/29)|. ve cac phUdng trinh bac ba hoac bac cao hdn nhung khong qua kho d i giai jirdc uhfnig phudng trinh do. ^,^ ^ ^ ^ / Vi du 1. Chon n = 2, f{x) = 2x - 3, b{x) = x + 1, a(x) = x - 1 to duac bai 01 3.3 Phifdng t r i n h [f{x)Y+h{x) = a{x) ^a{x).f{x) - b{x) todn sau. oc Bai toan 14. Giai phuong trinh iH 3.3.1 Phu'dng phap giai. Da 4x2 _ ^ + 10 - (x - 1) v/2x2 - 6 x + 2. Dat u = f{x) va v = ^a{x).f{x) - 6(x). ta thu dU0c he doi xiing loai I I (1) hi nT u" + b(x) = a{x).v 3- 3 +V5 Giai. Dieu kien x e -oo; U ; +00 . Taco •y" + b(x) = a{x).u. uO / ie Nh5n xet 1. Dang phucng trinh [/(x)]" + b{x) = a{x) ^a{x).f{x) - b{x){*) (1) (2x - 3)2 + (x + 1) = (x - 1) v / ( x - l ) ( 2 x - 3 ) - ( x + l ) . iL tSng qudt han dang phuang trinh {ax + 6)" = p ^f^T+V + qx + r da duac xet Ta d muc 3.2 d trang 160. PhUdng trinh trong bai toan 5 d trang 160 vik lai Dat u = 2x - 3, v= ^ ( x - l ) { 2 x - 3 ) - ( x + l ) . Ta c6 h$ s/ { 'u2 + x + 1 = ( x - 1 ) 7 ; (2) up (4x + 2 ) 2 - 1 4 = i y ^ i (4x + 2) + 14. t;2 + x + 1 = ( x - l ) u . (3) ro /g Vay no rcti vao dang phUdng trinh (*) Ung vdi om Lay (2) trit (3), ta dudc i;2 = ( x - l ) ( u - u ) fix) - 4x + 2, u = 2, b{x) = -14, a(x) = ^ . .c ^{u-v){u-\-v + x-l)^Q^ V =u ok V =I - X— u. PhUdng trinh trong bai todn 8 d trang 163 wiei /ai bo • Vdi r; = u, ta c6 ce (2x - 3 f - X+ 2 = 1^1- (2x - 3) + .T - 2. ? V ^ r - ? o . 2 ,0 o fa y2x2-6x + 2 = 2 x - 3 ^ ( Vgy no rdi vao dang phiCdng trinh (*) iing vdi \^ - 6x + 2 = 4x'' - 12x + 9 w. ^ { 2x2"-^6x + 7 = 0 (^^ "Shiem) . ww f{x) = 2x - 3, n = 3, 6(x) = - x + 2, a(x) = 1. • Vdi i; = 1—X - u, ta C O 3.3.2 Phifdng phap sang tac bai toan mdi. ^2x2 - 6x + 2 = 1 - X - (2x - 3)
V i d u 2 . Vdi n = 3, tic he doi xUng loai II, ta thu dicac ^ 3 ( x + l ) 2 - 8 x + 4 ^ 3x2 _ 2x + 7 ^ ^ ; u — V -v^ = {v - u) a{x) u"^ + v,v + + n{x) = 0. Yay t n r d n g hdp ^2 -|- uv + 7;2 + 1 _ 2x = 0 kliOng the xay ra. Phudng t r i u h da Ta se chon a{x),u{x),v{x) sao cho phUOng trinh + uv + -\- a{x) = Q ^lio CO duy nhat nghiem x = 1. > 1, A,, V uo nghiem. Ta co + uv + v' + a{x) — (^^ + 2 j 4 ' ^""^ 'fj g a i t o a n 16. Gid? phuang trinh 8x2 _ 13a: + 7 =: ^ 1 + ^ ) ^^3x2 - 2. (1) + 4a(x) > 0, c/io'n5 /lan chqn 3u^ + 4a{x) = Sx^ - 2x + 7 > 0, u = i + 1 khi do a{x) — 1 - 2x. Phuong trinh can sang tdc se la , t. G i a i . Dien kien x^O. PhUdng t r i n h (1) tUdng dUdng . ,.. / 01 (x + if + b{x) = (1 - 2x) ^ ( 1 - 2x) (x + 1) - 6(x). 8x^ - 13x2 + 7x = (x + 1) v/3x2 - 2 oc ^ (2x - 1)^ - (x2 - X - 1) - (x + 1) ^ ( x + 1) (2x - 1) + (x2 - X - 1). (2) iH Vay u = V se dan tdi Dat u = 2x - 1, v= s/(x + 1) (2x - 1) + (x2 - x - 1). K e t hdp vdi (2) t a c6 Da ^ ( 1 - 2x) (x + 1) - h{x) = X + 1
TT STT 77r V i d u 3. Tii vi du 2 d trang 170 ta c6 ^pjigni phan met 2cos—, 2 c o s — , 2 c o s — . (1 - 2 x ) (x + 1) - b{x) = (x+ if -b{x) = x^ + 5x2 ^ „ 2 , o / "\ 3u2 + 4 - 8x '-V'll I - ^ - > , u2 + ut; + T;2 + 1 - 2 x = (^1; + - j + Tic phuang trlnh Au^ - S-u = ^ 8u^ - 6u = 1, lay u= ta diwc ' ^ 3 (x + 1)2 - 8 x + 4 ^ 3x2 _ 2x + 7 ^ ^ x^ - 3 x = 1 - 6 ( x ) = 5x^ + 7x + 1. - 4 4 ' ' / Vay chon b{x) = - ( S x ^ + 7x + 1), khi do ' iraC! • = yay trirdng hop ^2 + +1;2 + i - 2x = 0 khong the xay ra. PhitOng t r i n h da / 01 cho CO ba nghiem phan biet x = 2 cos —, x = 2 cos — , x = 2 cos — . , [f{x)f + h{x) = a{x) ^ a ( x ) . / ( x ) - h{x) oc y y \j iH trd thanh Vf d u 4. Tii vi du2 d trang 170 ta c6 u • ;< • f ; Da (x + 1)^ - (5x2 ^ 7^ + 1) = (1 _ 2x) ^ ( 1 - 2 x ) (x + 1) + 5x2 + +7 (1 - 2x) (x + 1) - b{x) = (x + 1)^ ^ -b{x) = x^ + 5x2 ^ 4 ^ hi !i . X nT ^x^ - 2x2 - 4 x = (1 - 2^.) s / 3 x 2 + 6x + 2 fi?phuang trlnh Av? + 3u = 2, lay u = - , ta ditac uO - 6 ( x ) = 5 x 2 + x + 4. vi • ie iL Ta ri7r(/f i a i loan sau. Vay chon b{x) = - ( 5 x 2 + x + khi dd' ' • '' Ta „ , Ada viih ffi'Xfi'ji: B a i t o a n 17. Gidi philcfng trinh [f{x)f + 6(x) = a ( x ) ^ a ( x ) . / ( x ) - 6 ( x ) , s/ up x'^-2x-4 = (--2] \/3x2 + 6x + 2. (1) ro Vx y (x + 1)^ - (5x2 + a; + 4) = (1 _ 2x) y ( l - 2 x ) (x + 1) + 5x2 + x + 4 /g G i a i . Dieu kien x^O. (1) x^ - 2x2 _ 4^; ( 1 _ 2x) ^ 3 x 2 + 6x + 2, hay
• Vdi V = u ta. dUdc ,4.2 M o t so b a i t o a n r e n l u y e n v a n a n g cao. y/3x^ + 5 = X + 1 + 3x = 4. (3) ilii t o a n 19. Gidt phuong trinh 2{x'^ - 3.T + 2) = 3\/.r-* + 8. Diit X = 2u, t l i a y vao (3) t a duac Au^ + 3u = 2. (5iai. D i i u kien : x > - 3 . Phitdng trinh tUdng dUdng ' (4) V i ham so /(u) = 4^^ + 3u c6 /'(x) = 12^2 + 3 > 0, Vu e E nen phiTdng trinh 2(x^ - 2x + 4) - 2(x + 2) = 3^{x + 2){x^ - 2x + 4) ,, (4) CO khong qua m o t nghiem. Xet „ ^ a; + 2 x + 2 0. K h i do 2 - 2t^ = 3t ^ - 4T ). T a CO * - 2 Vav iH Do (16, iicu (lat a = '\/2+y/l thi 2 = ^ x^ ^- 2.7- + 4 t = -2(loai). Da 1 1 X 1 2 hi = 3 1 f 1\ + 4 - a = - 0. K h i do bo x^ + .r + 1 ce 3.4 Phu'dng trinh dSng cap doi vdi \fP{x) va v^Q(^) :;a' /jii.;'. fa 21',^ - 1 .+ ^ / = 0 2^/- + / - V 3 = 0 4=> w. Q P ( X ) + PQ{x) + Av/P(:r).Q(x) = 0 {a(3X ^ 0) t = p (k)ai) ww 2\/3 ^ • ^ 3.4.1 Phu'dng p h a p giai. x^ - X + 1 1 \l j^'x + l 75 '^^''^ - 4x + 2 = 0
{ x^ + X - 6 > 0 X - 1 >0 3^2 - 6.T + 19 > 0 ^ X >2. Phildng t r i n h tUdng dudrip x ( l + x ) = 1 - X 0 n e n c h i a c a h a i ve p h U d n g t r i n h x - 1 X - 1 \ om B a i t o a n 2 3 . Gidi •phUdng trinh / (1) cho (x2 + X + 1)2 > 0 t a d u d c 2-7. 13.- . Dat= ^x2 + X + 1^ x2 + X + 1 .c X - 1 t = -2 ok x' + 2x3 ^ 2x2 _2x^l= (x=* + x) sj . (1) t = . K h i d o 2 - 7
Quy (long bo rntiu ta duclc 2{x- + a; + 1) + 3(.T - 1) = 7\/{x - l)(.-r-' + x + Y yf du 3. Xet x = 2. Khi do Ta CO hdi todn sau [x^ + 2x + 2) = 10, X + 1 = 3,3(x^ + 2x + 2) - 8(x + 1) = 6, < B a i t o a n 25 ( D e n g h i O L Y P I C 3 0 / 0 4 / 2 0 0 7 ) . Gidi phuong Irmh (x + l)(x'^ + 2x + 2) = 30, (x + l)(x'^ + 2x + 2) = x^ + 3x'^ + 4x + 2. 2x + 6.x- - 1 = 7 V ^" - 1• ' (1 ~ yay x = 2 thi 3(x2 + 2x + 2) - 8(x + 1) = = %/x^ + 3x2 + + 2. G i a i . Dieu kien .T: > 1. f v30 v30 / (1) ^3(x- l ) + 2(x2 + x+ 1) = 7V(x- 1)(,T'-^ + X + 1). (2) fa CO bdi todn sau. ,» ,. • •- 01 g a i t o a n 27. Giai phuang trinh Sx^ - 2x - 2 = -^=\/x'^ + 3.r'2 + 4x + 2. oc V i X = 1 khong thoa (2) nen chia ca hai ve cua (2) cho x - 1 > 0 ta ditdc \/30 > iH ^3;2 + x + l ^ /x2+x + l G i a i . Dieu kien x^ + 3x2 + 4x -^ 2 > 0 (x + j)(^2 + 2x + 2) x > -1, Da phitdng t r i n h da cho viet lai hi nT 3(x2 + 2x + 2) - 8(x + 1) = ^ y ( x + 1)(T2 + 2x + 2). (1) Dat t = \ ^^'•^"'"^ =^ ^•• '^ + (1 - 0 ta iL \^2- 3 > 0 < = > ' > V 3 + 2V^. dUdc Ta s/ Phitdng t r i n h (3) t r d t h a n h 2 i - - 7^ + 3 = 0 f G {'^' ^""^"^ '''^ x+ l v/30V x+ l ^' up kien ciia, t t a ditdc \ 3. Vay Dat t= + 2^ + 2 > 0. K h i do ro - 1 /g 3(2 - 8 = - J = ( 4=^ 3t2 - - ^ t - 8 = 0 3v/30(2 - 6i - 8^30 = 0. (3) om X v/30 v/30 .c Ket hdp v d i d i e u k i e n t a dUdC' x = 4 ± %/6 l a tat ca cac nghiem cua (1). ok L i f u y. Goi g ( x ) = X - 1, P ( x ) = + X + 1. M a n c h o t c i i a Idi giai l?i p i i a i i Nhan xet rang i la nghiem ditdng ciia phiTdng t r i n h (3), hay t = yj^- Vay t i c h v6 t r a i c i i a (1) t h a n h : VT = 2P{x) + SQ{x). T i n h y t a se thRy 2 l a lu' bo s o c i i a x^ t r o n g v c t r a i c i i a (1). Cung tir d o suy ra 3. T u y n h i e n de d a n g t i m no 9 x = 2^ ce x2 + 2x + 2 d u d c c a c so 2 v a 3 b a n g p h U d n g p h a p he so b a t dinh —-—= J—
3.5.1 P h L f d n g p h a p g i a i v a m o t so b a i t o a n r e n l u y e n , n a n g ca.; B a i t o a n 30. Cho ham so f{x) = v ^ T T x + - x + v ' ( l + x)(8 - x). a) Tim gid tri Idn nhdt, gid tri nho nhdt, tap gid tri cua ham s6 ciia ham so. Dat t = y/P(x) ± s/Q(x). K h i do = P{x) + Q{x) ±2^P{T)Q{X). T h a y v,V, b) Gidi phuang trinh \ / l + x + V8 - x + ^/{YTlc){S^^ = 3. phUdng t r i n h da cho t a dttdc + pt + \ 0. c) Tim rn de phuang trinh y/l + x+y^S - x + ^ / U + x)(8 - x) = ni c6 nghiem. B a i t o a n 2 8 ( D H - 2 0 1 1 B , p h a n c h u n g ) . Giai phuang trinh (],) Tim rn. de phuang trinh s/l + x+\/8 - x-\-^J{l + x)(8 - x ) = m cd nghiem duy nhdt. 3 ^ 2 1 ^ - 6 \ / 2 ^ + 4 ^ 4 - x2 = 10 - 3x- (x e E ) . (]> Giai. a) Tap xac d i n h cua ham so la [ - 1 ; 8]. T a c6 / G i a i . Dieu kien - 2 < x < 2. Dat t = 3^2 + x - 6\/2 - x. K h i do ' 01 1 1 7-2x 1 oc f2 = 9 (2 + x ) - 36v/4 - x2 + 36 (2 - x ) = 9 (lO - 3x - 4v/4 - x2) . 2^/1+^ 2^/8^^ 2v/(l+a;)(8-x) it iH T' V ^ [ £ x - _ v / r T x _^ _ _ ^ ^ ^ 2 ^ _ _ Da t = 2v/l + x . v ^ 8 ^ 2v^(l + x ) ( 8 - x ) ;:t sat' Thay vao (1) t a diWc t = — ^t^-9t = 0
,/ = 0 3, dau bang xay ra k h i x = - 1 . T o m lai Suy ra Gx + 4s/2x'^ + 5x + 3 = 2f2 - 8. Thay vao (1) t a dUdc 7 fa t = 5 w. gia t r i Idn nhat ciia u la 3\/2, dat dirdc k h i x = - ; gia t r i nho nhat ciia u la 3i = 2 ^'P'^' xx — b-a-x. Thay ^ { i 2 ^ - ' ' l 4 6 x + 429 = 0 ^ { X = 143 v;^:': X = ——^ vdo (*) di ttm m, sau do thiJC lai. X = 3 la nghiem d u y nhat ciia phudng t r i n h d a cho. . 182 183
3.6 M o t so hijfoing sang t a c phufdng t r i n h vo t i . I
V i d u 3. Cho t o a n 38. Gidi phuong trinh a = v/3x2 - a;+ 2001, 6 = - v^Sx^ -lx + 2002, c = - v^Gi - 2003 Vx^ + 3x + 2 ( N / T T T - v ^ T 2 ) = 1 . r : , ? thi + 6^ + = 2002. Ta dUdc bai toan sau. ,, G i a i . Tap xac d i i i l i R . PhUdng trinh viet lai .T •vt.nf^ P I • B a i t o a n 35. Gidi phUdng trinh ( x + 1) + (x - 2) + \/x2 + 3x + 2 ( ^ x + 1 - v^x + 2) = 0. (2) \/3.T2 - X + 2001 - v/3.r2 _ 73. + 2002 - ^^^6.7; - 2003 = v'2002. Pat a = \/x + 1, 6 = - v/x + 2. T h a y van (2) t a diWr , v / Hu-dng d a n . Dat 01 a^ + b^- ah{a + 6) = 0 (a + 6)(a - bf = 0
• Khi u = \/2?;, t a dUdc y 4. Viec tao ra cdc dang thiic tUdng tu nhu t 31/;;:? 1 v/.r + 2 = \/2v'.r2 - 2.T + 4
r Giai. Dicu kicn x> - - . PhUdng trinh viet hii ^^i toan 42. Gidi. phMng trinh 2x^+4x= «o r (3x + 2)2 - 6 = N/3"X + 8. (1^ y j du 11- C^ho CY = 2, = - 1 , a = 8000, h=l thay vao (*) ta duac Dat 3?y + 2 = \/3x + 8, suy ra (3?; + 2f = 3.r + 8. Ket h0p vdi (1) ta c6 he (2x - 1)2 = 4000\/8000x- + 1 + 4001. '' (3x + 2)2 = 3y + 8 (2) CO bai toan sau. (3t/+ 2)2 = 3x + 8. (3) g^i toan 43. Gtdi phuang trinh x2 - x - 1000\/8000x + 1 = 1000. / '"' 8 8 01 KT^u xet he [ i " ^ oc De x,y thoa man (1) va (2) thi x > - - va y > - - . Lay (2) tiif (3), ta duuc + ? thi t i l phUdng triiih dudi, ta dUdc ' • t [ay + p) = ax +1) iH •,,, 3(xX -- j/)(3x y =- Xx) ^ (x - y)(3x + 3y + 5) = 0 y = 0 + 3y + 4)^- 3(y Da 3x + 3y + 5 = 0 ^ 3 y = -(3x + 5). ay + /3 = s/ax + 6
V i d u 14. Xet ham so f{t) = + 2t dong bien tren R. Cho jjifdlng dan. Bai toan nay chinh la bai toan 12 d trang 125. V i d u 16. Xet ham so f{t) = t^ + t dong bien tren R, roi xet phuong trinh f (^-x'3 + 9 x 2 - 1 9 x - + l l ) = f{x - 1). f{x + 1) = fW^x + 1), ta diCac bai toan sau. ta dicac ' ' gai toan 48. Gidi phuang trinh x^ + 3x2 j^^x + 2 = (3x + 2)V3x + 1. (1). _j-3 + _ + 11 + 2 v / - x 3 + 9 x 2 - 1 9 x + l l = (x - 1)^ + 2{x - 1). Giai. Dieu kien x > Khai triin vd rut gon ta dUcic bai loan sau. / 01 (x + l ) 3 + x + l = ( 3 x + l + l ) v / 3 7 n '^f' B a i t o a n 46 (De n g h i O L Y M P I C 3 0 / 0 4 / 2 0 0 9 ) . Gidi phUdng trinh oc -^(x + 1)3 + X 4-1 = V3x + 1^ + \/3x + 1 ^f[x + 1) = /(V37TT) vdi / ( i ) = r. ,, , iH - 6.T2 + 12.T - 7 = \ / - . r 3 + 9 . T 2 - 19x+ll. ^t Da G i a i . Dat y = sj-x^ + ^x^ - 19x + 11. Ta c6 he ^ x + 1 = V3x + 1 (do J{t) = 1? + t dong bien) y; hi „ 3 ^ _3,3+ 93.2 _ 19^ + 11 ^ / y3 = + 9x2 - 19a; + 11 0. Ta c6 he f (x + 1)^ + (x + 1) = (7x2 ^ 93. _ 4) + A/7X2 + 9x - 4 r 2x3 + 7x2 + 5x + 4 = 2y3
2 trong fit) t a bien d5i ve t r a i ciia' (2) thanh : (2x + l){V4x^ + 4x + 4 4- 2). Xet ham so f{t) = 2t^+t'^. V i /'(t) = 6f + 2t > 0 , V i > 0 nen ham so / dfing pg thay 4x^ + 4x + 4 = (2x + l ) ^ + 3. Vay t a da xay dung thanh cong h a m bien tren [0; +00). Lfii c6 y > 0, tit dieu kien suy ra a; + 1 > 0, do do d5ng bien f{t) = t (2 + Vt^ + 3 ) . so (*) f{x + 1) = /(y) < ^ x + l = y < ^ x + l = VSx - 1 • i< V i d u 2 0 . Xet ham so /(
Xet ham so /(t) = logg t - 2t + yt > 0. Ta c6 Xet ham so g{x) = \/4x + 5 4- \/x + 8,Vx > - - . V i 1 fit) = + 2t - 2 > 2 :.2i-2 = 2 - 2> 0