Tài liệu Onluyentheocautrucdethimontoan phan02(2)

CHI DAN Xet dau tarn thuTc h(x) = 2x^ - 3x - 5, c6 h(x) = Oc^x = - l , x = - 2 2 5 5 h(x) > 0 vdi X < - 1 hoac x > — ; h(x) < 0 vdi - 1 < x < — Xet tLfcmg tir vdi tam thufc g(x) = 4x^ - 19x + 12 ta lap diigfc bang sau : X + ^ _ h(x) g(x) + h(x) = 2x' - 3x - 5 -1 + g(x) = 4x2 - 19x + 12 0 - J r 5 2 - - + - 0 0 ^ + - + 0 0 + 0 - + + TCf bang xet dau ta ket luan : f(x) > 0 vdi X e (-oo; -1) u ' 3 . v4' 5^ u (4; 2y +oo) rs ^ 3^ f(x) < 0 vdi X e f(n) = 0 vdi X = - 1 , X = 2 3 f(n) khong xac dinh vdi n = —, x = 4. 4 0. 18.a) 2x2 - bat + 2 > 0 t r i n h va bieu d ib) x^ + nghiem< tren true so. Giai 5x phifong l n tap 4x + 3 HI DAN a) Xet dau tam thufc ve trai : fix) = 2x2 _ 5x + 2 A = 52 - 16 > 0, fix) = 0 Xj = - = 2 , X2 fix) > 0 (f(x) cung dau vdi 2) vdi x < - hoac x > 2. Vay tap nghiem 2 1] 2; + c o ) . Bieu dien tap cua bat phirong t r i n h la: T = { -00; — I 2_ nghiem tren true so' (phan true so khong bi danh cheo). 1. ^ ] » « f ^ • b) Lam nhiT cau a). Tap nghiem T = (-3; -1). -3 -1 32 S TS. Vu The Hi^u - Nguygn VTnh CSn 19. Giai he bat phifofng trinh va bieu dien tap nghiem tren true so : (I) x' + x - 6 > 0 (1) x'-5x +4<0 (2) CHI D A N Tap nghiem cua (1) : T i = (-oo; - 3 ] u [2; +oo). Tap nghiem ciia (2) : T2 = (1; 4). -3 (D(2)i (i)i Tap nghiem cua he (I) : T = T i n T2 = [2; 4). ^ 20. Trong moi trirdng hop, t i m cac gia t r i tham so m de : a) PhtfOng t r i n h : (3m + l)x^ - (3m + l ) x + m + 4 = 0 (1) c6 hai nghiem phan biet. b) Phuong trinh : (m - 4)x^ + (m + l ) x + 2m - 1 = 0 (2) v6 nghiem. CHI D A N a) PhLfOng t r i n h (1) c6 hai nghiem neu 3m + 1 ;t 0 va biet thuTc m ^ — 3 -.(3m + l)(m + 15) > 0 Tarn thufc theo bien m : A(m) = -3m^ - 46m - 15 c6 cac nghiem : m i = -15, A = (2m + - 4(m + 4)(3m + 1) > 0 < m2 = - — CO gia t r i diiong (trai dau vdi - 3 ) vdi -15 < m < - —. 3 3 Ket luan : Phirong t r i n h (1) c6 hai nghiem neu -15 < m < - - . b) Neu m = 4 phirong trinh (2) c6 dang : 5x + 7 = 0 c6 mot nghiem. . fm ^ 4 PhUOng trinh (2) v6 nghiem neu \ ^ ' ^ • [A = (m + l ) ' - 4 ( m - 4 ) ( 2 m - l ) < 0 A = -7m^ + 38m - 15 c6 gia t r i am (cung dau vdi - 7 ) neu m lay gia t r i 3 ngoai khoang hai nghiem la : m.^ ^ — va m2 = 5. 21. DS : m < — hoac m > 5. 7 Tim cac gia t r i cua tham so m de bat phufong t r i n h x^ - (3m - i)x + 3m - 2 > 0 (1) dutJc nghiem dung vdi moi x thoa man I x I > 2. CHI DAN Tarn thufc fix) = x^ - (3m - l)x + 3m - 2 c6 biet thufc A = (3m - 1) - 4(3m - 2) = 9(m - 1)'. • Neu m = 1, A = 0=:> f(x) > 0 Vx e R, f(x) = 0 o x = 1, do do. f(x) > 0 Vx: > 2. Hoc 6n luy?n theo CTDT mfln Todn THPT S 33 22. Neu m vi 1 => f ( x ) = 0 < > x i = 1, x = 3 m - 2. = Vdfi m < 1 t h i 3 m - 2 < 1 tap nghiem cua (1) la: ( - o o ; 3m - 2) u ( 1 ; +oo) De X sao cho x | > 2 thuoc t a p n g h i e m t h i p h a i c6 - 2 < 3 m - 2 c^O < m < 1. Vdfi m > 1 t h i 3m - 2 > 1 tap nghiem cua (1) la: ( - o o ; 1) u (3m - 2; +oo) De t h o a m a n yeu cau b a i t o a n t h i p h a i c6 3m-2<2<=>l 2 thuoc t a p n g h i e m cua (1) t h i m p h a i 4 lay gia t r i sao cho : 0 < m < T r o n g m o i triTdng hop diidi day, h a y t i m cac gia t r i cua t h a m so m de a) phirofng t r i n h : (m - l)x^ - 2mx + m + 5 = 0 (1) CO m o t n g h i e m thuoc k h o a n g (0; 1). b) phirong t r i n h : 4x^ - (3m + l ) x - m - 2 = 0 (2) CO h a i n g h i e m thuoc k h o a n g ( - 1 ; 2). CHI DAN a) Neu m = 1, phtTong t r i n h (1) c6 n g h i e m duy n h a t x = 3 g (0; 1). Neu m ^ 1, phirong t r i n h (1) la phiiong t r i n h bac h a i . A p dung d i n h l i dao ve dau cua t a m thiJc, phiTong t r i n h fix) = (m - l)x^ - 2mx + m + 5 = 0 CO m o t n g h i e m t r o n g khoang (0; 1) neu va chi neu fIO).fll) < 0 o (m + 5)4 < 0 o m < -5. b) PhiTcfng t r i n h (2) c6 h a i n g h i e m t r o n g khoang ( - 1 ; 2) neu m lay cdc gia t r i thoa m a n cac dieu k i e n sau : A = (3m + + 16(m + 2) > 0 af (-1) = 4[4 + (3m + 1) - m - 2] > 0 af (2) = 4[16 - 2(3m + 1) - m - 2] > 0 , S 3m + l -1 < —= < 2 2 8 -3 < m < 5 - 8 < 3 m + 1 < 16 3 m > — 2 12 m < — - 7 m + 12 > 0 9 m ' + 2 m + 33 > 0 2m + 3 > 0 23. 3 12 <=> — < m < — . 2 7 T i m cac gia t r i cua t h a m so m de phirong t r i n h (m - 2)x^ - 2mx + 2 m - 3 = 0 (1) CO h a i n g h i e m thoa m a n dieu k i e n - 6 < X i < 4 < X 2 . £ 2 TS. Vy Th6' H^fu - NguySn Vinh CJn 34 CHI DAN - 6 < x i < 4 < X2 o af (4) = (m - 2)(10m - 35) < 0 af (-6) = (m - 2)(50m - 75) > 0 c>2 2 t h i (1) c6 hai nghiem thoa man xi < 2 < X2. • Neu 2 - m = 0 < > m = 2 t h i (1) c6 cac nghiem 1 = x i < X2 = 2. = • Neu 2 - m > 0 o m < 2 t a xet them biet thufc A = (m + 1)^ - 4m = (m - 1)^ S „ m+1 „ m-3 o\ 2= 2= < 0 (vi m < 2) 2 2 2 Vdi m = 1 ta CO nghiem kep x i = X2 = 1 < 2 Vdi m < 2 va m ^ 1 t h i X i < X 2 < 2. 25. Giai, bien luan theo tham so m he bat phiTOng t r i n h (I) x ' - (m + l)x + 2(m - 1) < 0 (1) x' - (m + 2)x + 3(m - 1) > 0 (2) CHI DAN Tam thu-c ve trai (1) la : fix) = x^ - (m + l ) x + 2(m - 1) = 0 o X = m - 1 hoac x = 2 • Neu m - 1 < 2 o m < 3 t h i (1) c6 tap nghiem • Neu m - 1 > 2 Tj = [m - 1; 2]. m > 3 t h i nghiem cua (1) la : Ti = [2; m - 1]. • Neu m = 3 t h i T i = 121. Xet tam thufc ve trai cua bat phtTcfng trinh (2) : g(x) = x^ - (m + 2)x + 3(m - 1) c6 cac nghiem x = m - l v a x = 3. • Neu m < 4 tap nghiem cua (2) la : T2 = (-00; m - 1] u [3; +co). • Neu m > 4 tap nghiem cua (2) la : T2 = ( - c o ; 3] u [m - 1; +00). • Neu m = 4 : T2 = {3}. Tap nghiem T = T i n T2 cua he (I) nhif sau : + Neu m < 3 t h i (I) c6 nghiem chung duy nhat x = m - 1. + Neu 3 < m < 4 t h i T = [2; m - 1]. + Neu m = 4 t h i (I) c6 hai nghiem chung T = {2; 3}. + Neu m > 4 t h i T = [2; 3] u !m - 1}. HQC va 6n luy?n theo CTDT m6n Toan THPT S 35 S 1. §4. PHlIC(NG T R I N H , B A T P H U C I N G T R I N H CHlfA D A U GIA TRI T U Y E T DOI KIEN THLfC Phx:ifofng t r i n h chtfa d a u g i a t r i t u y ^ t d o i La phucfng t r i n h t r o n g bieu thufc cua no c6 chufa a n n a m t r o n g dau gia t r i tuyet do'i. Vt : I 2x - 1 1 - 3x = 2. Dudng l o i g i a i phirong t r i n h a neu l a diTa t r e n d i n h nghia a > 0 -a neu a < 0 ta phan chia tap xac d i n h t h a n h nhufng tap nho de thay the bieu thufc c6 gia t r i tuyet doi bkng bieu thufc khong c6 gia t r i tuyet doi ti/cfng diTOng. Dac b i e t : | A(x) | = | B(x) i o (A(x)f = (B(x)f'. 2. B a t phu:cifng t r i n h chtfa d a u g i a t r i t u y $ t d o i Cac phtrong t r i n h don g i a n chufa dau gia t r i tuyet doi : M o t so t i n h chat cija gia t r i tuyet doi can n h d k h i g i a i bat phuong t r i n h chufa dau gia t r i tuyet doi. 2. a.b 1. I fix) I < g(x) <=> - g ( x ) < fix) < g(x). c) I f(x) I > g(x) b) I fix) I < I g(x) I « a) 3. o f ' ( x ) < g'(x). fix) < - g ( x ) hoSc fix) > g(x). -a a 4. a + b + BAI TAP 26. G i a i cac phuong t r i n h : - 2x = 7 (1) b) X + 4 a CHI DAN X + a) Theo d i n h nghia I x + 4 (A) (1) « (B) . 4 2x + 3 vdi - ( x + 4) v d i X > X < 3x - 2 = 3 (2) -4 -4 Jlx + 4 ) - 2 x = 7 X > -4 - ( x + 4) - 2x = 7 -4 X < 63 TS. Vu Thg Hi^u - Nguyln VFnh CJn 36 -x + 4 = 7 _ (B) o <=>x = - 3 ; r-3x-4 =7 = > v 6 nghiem X < -4 -4 Vay phuong t r i n h (1) c6 nghiem duy nhat x = - 3 . b) De CO bieu thufc khong chufa gia t r i tuyet doi ttrcrng duong ta lap bang sau (A) < X > 3 "2 X 2 3 2x + 3 -2x - 3 0 2x + 3 2x + 3 3x- 2 -3x + 2 Bieu thufc (2) -5x - 1 = 3 - X Nghiem X = — (loai) 5 X = 2 (loai) -3x + 2 0 3x - 2 +5=3 5x + 1 = 3 X = — (loai) 5 Vay phiiong t r i n h (2) v6 nghiem (theo bang c6 nghia la vdi x < — 2 4 thi (2) CO bieu thufc : - 5 x - l = 3 = > x = khong thuoc khoang 3 nen loai). 5 -2 27. Tuong tir nhir vay, ta loai x = 2 va x = —. 5 Giaix cacxphiTOng t r i n h : ^2x-4 = 3 ( 1 ) b ) 2 x ' + 6x + 8 + x^ - 1 a) =30(2) CHI DAN a) x ^ - x = 0 « x = 0; x = l , 2 x - 4 = 0 o x = 2 x^ - X x^-x -x^ + x vdi X < 0 vdi 0 < 2x - 4 vdi 2x- 4 -2x + 4 vdi X < X > x >1 1 2 X < hoac 2 Ta lap bang de d i theo doi. X x^-x 2x- 4 0 -00 x^ - 0 X -2x + 4 1 -x^ + -2x + 4 Bieu thufc (1) x^ - 3x + 4 = 3 -x^ Nghiem 3±V5 (loai) 2 X X + 2 0 x^ - x^-x X -2x + 4 +00 0 2x - 4 4 = 3 x^ - 3x + 4 = 3 x^ + - 1 +Vs 2 3±V5 — - — (loai) Hoc X - 4=3 - 1 + V29 2 6n luyOn theo CTDT mSn Toan THPT S 37 Phifdng t r i n h (1) c6 cac nghiem : X j = - i + Vs. _ 1 _ - 1 + V29 X 9 b) Ta lap bang sau : X -00 -4 -2{x^ + 6x + 8) 2|x^ + ex + 8l -2 x'^- 1 Ix^-ll 2(x2 + 6 x + 8) 0 B i e u thiifc (2) 3x^ + 12x + 15 = 30 -1 1 0 2(x^ + Gx + 8 x^ - 1 -x2-12x-17=30 2(x^ + 6x + g 0 -x^ + 1 (c) +CO 2(x^ + 6x + 8) 0 x^ - 1 (d) (e). Nghiem (c) o _ ^2x + 15 = 30 ^ (d) x^ + 12x + 17 = 30 (e) 3x2 _^ -^2x + 15 = 30 TCr cac phirong t r i n h trong cac khoang ta t i m dtfoc cac nghiem cQa (2) la : X = - 5 , X = 1. Giai cac phirong t r i n h : 2x-l -3x = 1 b) 2x - 3 X + 4 =6 28. a) c) x 2 - x = 3 x - 1 CHI DAN 2 b) X = 1 va X = — c) X = 1, X = ±3. a) X = 3 11 29. Giai cac bat phuong t r i n h va bieu dien tap nghiem tren true so. a) | 3 x - l | < 2 (1) b) i'2x+*l| > 3 ' (2) CHI DAN a) (1) <;> - 2 < 3x - 1 < 2 <=> - 1 < 3x < 3 « — 3 < X < 2x + l > 3 1 p3 1. 2x + 1 < - 3 b) (2) •/W/MW/////////////, <: => x < -2 -2 1 x>l Giai bat phuong t r i n h va bieu dien tap nghiem tren true so'. _|x2 - 2 x - 3 f < 3 x - 3 (1) 30. CHI DAN (1) o-(3x-3)<(x2-2x-3)<3x-3 rx<-3 0 fx' + X - 6 > 0 < X < 5 x>2 x ' - 5x < 0 0 < X < <=> 2 < X < 5. 5 3 8 C3 TS. Vu Th6' Huu - Nguyin VTnh CJn §5. PHlIOfNG TRINH, BAT PHlJOfNG TRINH CHlfA CAN THtJC KIEN THCfC 1. PhvfoTng t r i n h chufa c a n thijfc PhiTofng phap chung de g i a i phtTOng t r i n h c6 bieu thufc chufa a n nam diTdi dau can l a n a n g l e n l u y thCra bieu thufc cua phifcfng t r i n h v d i cac dieu k i e n d i k e m de c6 diroc phucfng t r i n h k h o n g con chufa a n t r o n g dau can. Cung c6 the dSt a n p h u de d a n d e n cac phifcfng t r i n h , he phuong t r i n h don g i a n va de g i a i h o n . M o t v a i dang phifOng t r i n h chufa can thiJc dan g i a n l a : Jg(x) > 0 a) VfOO=g(x) ^ |f (x) = (g(x))' f (x) = g(x) b) V f U ) = Vg(x) 2. [f(x)>0, g(x)>0' B a t phu!ofng t r i n h chufa c a n thiJc Dang CO b a n cua b a t phiTOng t r i n h chura can bac h a i . f (x) > 0 a) g(x) < 0 4Ux)>gix) g(x) > 0 • f(x) > [g(x)]^ f (x) > 0 b) V f O O < g ( x ) ojg(x)>0 f (x) < [ g ( x ) ] ' De g i a i cac b a t phiTOng t r i n h chufa cSn thufc, t a dtfa r a cac dieu k i e n xac d i n h r o i luy thCra m o t each t h i c h hop cac ve cua b a t phiTOng t r i n h de g i a m d a n cac dau can thufc, d a n d a n dtTa t d i b a t phifong t r i n h , he bat phufong t r i n h k h o n g chuTa cSn thufc. Cung c6 t h e dat cac a n p h u hoSc b i e n l u a n cac ve cua b a t phuong t r i n h de t i m n g h i e m . BAITAP 31. a) G i a i cac phiTOng trinh : x - V2x + 7 = 4 (1) b) V3x + 4 - V x - 3 = 3. CHI D A N a) (1) V2x + 7 = X - 4 x-4 >0 2x + 7 - ( x - 4 f Hgc vJ On luy§n theo CTBT m6n To^n THPT S 39 <=> < X > 4 x ' - lOx + 9 = 0 b) ( 2 ) o V3x + 4 = 3 + V x - 3 c ^ o 9(x - 3) = (x - If 32. X > 4 -irx = 1 C:>X = 9. 3Vx-3 = x - 1 3x + 4 = 6 + x + 6 V x - 3 x>3 x-3>0 o x^ - l l x + 28 = 0 <^ x =4 x = 7' Giai cac phuang t r i n h : a) 2x - x^ + V e x ' - 12x + 7 = 0 (1) b) V x ' - 3x + 3 + V x ' - 3x + 6 = 3 (2) CHI DAN a) Nhan x e t : 6x^ - 12x + 7 = 6(x - 1)^ + 1 > 0 Vx e R. Dat t = V e x ' - 1 2 x + 7 vdi t > 0, ta c6 : x^ - 2x = 7-t' 6 va phirang t r i n h (1) dan den t = -1 +t =0 (loai) t =7 o Vex' - 12x + 7 = t > 0 6x' - 12x + 7 = t ' < > ex^ - 12x + 7 = 49 c:> xi,2 = 1 ± V s . = b) Nhan xet : x^ - 3x + 3 = Dat t = x^ - 3x + 3, ta 3 \ + - > 0 Vx 4 X 2 CO : G R. t ' + 3t = 9 + 1 ' - et V t ' + 3t = 3 - t t >0 t >0 2t + 3 + 2Vt(t + 3) = 9 Vt + V t T 3 = 3 (2) 0 x ^ - 3 x + 2 = 0 o x = l hoac x = 2. Giai phiTcfng t r i n h : Vx - 2 + V4 - x = x ' - 6x + 11 (11) CHI DAN Ve phai (1) la : x^ - 6x + 11 = (x - 3)^ + 2 > 2 Vx Theo bat dang thiJc Bunhiacopxki ve t r a i V x - 2 + V 4 - X = l . V x - 2 + 1.V4 - X 0 Do do : (1) 4-x> 0 N/X-2 + V4-X 34. Giai cac phuong t r i n h : a) ^ / ^ 7 3 4 - ^ / I ^ = l (1) CHI = 3. ox x' - 6 x + l l = 2 =2 b) ^/^r75+^x + 6- = fe + l l (2) DAN a) Cdc/i i : Lap phUdng hai ve (1) ta difoc : ( D o (x + 34) - (x - 3) - 3^x + 34.^/^r^[^x + 34 - ^ x - 3 ] = 1 (Ap dung hang dang thuTc : (a - b)^ = a^ - b^ - 3ab(a - b)) ( D o ^(x + 34)(x - 3) = 12 o (x + 33)(x - 3) = 1728 "x = 30 o x ^ + 31x - 1830 = 0 o x = -6l' Cdch 2 : DSt an so phu u = ^x + 34, v = ^ x - 3 ta c6 : (1) o u - V = 1, u^ = 37 1 = (u - v)^ = u^ - 3uv(u - v) o 1 = 37 - 3uv => uv = 12 u + (-v) = 1 Ta dilgfc he phirong t r i n h u ( - v ) = -12 u va - V la nghiem cua phifcfng t r i n h - X - 12 = 0 o Xi = - 3 , X2 = 4 ^x + 34 = 4 '^x + 34 = -3 x = 30 hoac o X--61' ^/^r^ = 3 ^/^r^ = -4 b) Dat u = ^x + 5, V = ^x + 6 ta thay u^ + v^ = 2x + 11. Suy ra (u + v f = + v^ + 3uv(u + v) => 2x + 11 = 2x + 11 + 3uv(u + v) o uv(u + v) = 0 o ^x + 5.^x + 6.^2x + l l = 0 =0 o ^x + 6 = 0 ^2x + l l = 0 35. x = -5 o x = -6 X 11 = -- Giai, bien luan theo tham so m cac phifOng t r i n h : a) Vx + m + V x - m = V2m (1) CHI b) Vx^ - 2mx + 1 + 2 = m (2) DAN a) DK : m > 0. • Ne'u m = 0 t h i (1) c6 nghiem x = 0. X > m >0 • Neu m > 0(1) o X 2x + 2Vx^ - m ^ = 2m o HQC > m >0 4^ m =m- X 6n luygn theo C T D T mOn Toan T H P T 0 41 X > m <=>m-x>0 o x =m x^-m'=(m-x)' Ket luan : V d i m > 0, phuong t r i n h (1) luon c6 mot nghiem x = m . b) TCr bieu thiJc cua phirong t r i n h t a t h a y dieu k i e n xac d i n h m > 0. m-2>0 (2) o Vx^ - 2mx + 1 = m - 2 x ' - 2mx + 1 = ( m - 2)2 m > 2 | ( x - m ) ' - 2m^-4m +3 (*) V i 2m^ - 4 m + 3 = 2(m - 1)^ + 1 > 0 V m n e n phirong t r i n h (*) luon c6 36. 2 n g h i e m p h a n b i e t xi,2 = m ± V2m^ - 4 m Ket luan : N e u m < 2 t h i (2) v6 n g h i e m , 2 nghiem. V d i cac gia t r i nao cua t h a m so m phifcfng V3 + X + V6 - X - V(3 + x)(6 - x) = m +3 n e u m > 2 phiTofng t r i n h c6 t r i n h c6 n g h i e m . (1) CHI DAN Cdch 1 : D K X D : - 3 < x < 6. DSt u = VSHOC > 0,V = V6 - x > 0. K h i do (1) <=> u + V - uv = m (2) (2) => uv = u + V - m (u + v ^2 = ..2' + . 2 + 2uv = (3 + x ) + (6 - x ) + 2uv )' u . = 9 + 2uv = 9 + 2(u + V - m ) (u + v)^ - 2(u + v ) + 2m - 9 = 0 u +V = 1+Vl0-2m ^ ju^ + (loai u + v = 1 - V l 0 - 2 m ) =9 Ta t h a y u + v = Vu^ + + 2uv > Vu^ + = 3 ( v i uv > 0) Mat khac, theo b a t dSng thufc Co si - Svac u + V = l . u + l . v < V ( l ' + l ' ) ( u ' + v ' ) - 3V2 V a y 3 < u + v < 3 V 2 ^ 3 < 1 + VlO - 2m < 3V2 ~ ^< m < 3 2 6V2 — 9 Ket luan : Ydi < m < 3 phiTcfng t r i n h (1) c6 n g h i e m . 2 Cdch 2 : D u n g phiTOng phap khao sat h a m so. Dat t - V3 + X + V 6 - X vdri - 3 < x < 6. T i m gia t r i \dn n h a t , gia t r i n h oV6 - txcua t +rXn [^ 3 ; 6 ] . T a c6 : n h a - V3 t e 3 t'(x) = —, =—, ,—=^ = 0 o X =— 1 2V3 + X 2 V 1 - X 6 2V3 + X . V 6 - X 2 M i n t(x) = m i n - | t ( - 3 ) ; t - ; t(6)^ = min{3; 3^2; 3} = 3 v2y 42 ^ TS. VQ ThS' H^u - Nguygn VTnh C?n Max t = 3V2 Vay t e [3; 3>/2] Ta c6: = (3 + x) + (6 - x) + 2^(3 + x)(6 - x) => V(3 + x)(6 - x) = Ve trai cua phiTOng t r i n h (1) t r d thanh : ^t^-9^ = - - t ^ + t + - = m v d i t e [3; 3V2] fit) = t 2 2 Khao sat SLT bien thien cua fit) tren [3; 3V2] ta thay : f'(t) = - t + 1 < 0 Vt t [3; 3V2] G ^ f t 3yf2 6\f2i — 9 TCf bang bien thien cho thay : < m < 3 t h i phirong t r i n h (1) C O nghiem. 3 7 . Giai cac phuong t r i n h : b) V3x + 4 - V x - 3 = 3 a) N / X - 1 = 13 d)>/5x-l = V 3 x - 2 - V 2 x - 3 . c) ^Jl6-x + ^|9 + x= ^ X CHI + D A N a) = 10 c) X = 0, x = 7 b) 4 va = 7 d) v6 nghiem. X 38. a) X = X Vdi gia t r i nao cua tham so m phiiOng trinh diTdi day c6 nghiem : X + m = 2Vx + 3 b) V x - 1 + V 3 - x - V(x - 1)(3 - x) = m C H I D A N a) m < 4 3 9 . Giai cac phiiOng t r i n h : b) 1 < m < V2. a) 2 ^ 3 x - 2 + 3 V 6 - 5 x - 8 = 0 (1) (Trich de thi tuyen sinh DH khoi A - 2009) b) 3V2 + X - 6V2 - X + 4N/4 - X ' = 10 - 3 X (2) (Trich de thi tuyen sinh DH khoi B - 2011) C H I D A N a) DKXD : 6 - 5x > 0 o he phifcfng t r i n h : x < - . DSt u = ^3x - 2, v = N / 6 - 5 X , f2u.3v = 8 5u^ + 3v^ = 8 ^ V V > 0 ta C6 = 15u' + 4u' - 32u + 40 = 0 Hoc va 6n luy?n theo CTBT m6n ToSn THPT S 43 b) V = 8-2u <=> <=> i (u + 2 ) ( 1 5 u ' - 2 6 u + 20) = 0 ru = V = -2 4 =^ X = -2. D K X D : - 2 < X < 2 . D a t u = V2 + x , v = V2 - x t a c6 h e : (I) + = 4 3u - 6 v + 4 u v = 3v^ + 4 Tii (*) => 3 ( u - 2 v ) = ( u + = 4 u - 2v = 0 ( D o + = 4 u - 2v = 3 2vf 3 u - 6 v = 4v' - 4 u v + u'(*) u - 2v = 0 u - 2v = 3 (A) (B) H e ( A ) CO n g h i e m v ^ = — o 5 40. 2-x = — =>x 5 = —. 5 H e ( B ) v 6 n g h i e m . V a y phucfng t r i n h (2) c6 n g h i e m d u y n h a t x = - . 5 G i a i cac phtrcfng t r i n h : a) 2Vx + 2 + 2>/^7T - Vx + 1 = 4 b) (1) (Trich de thi tuyen (2) V2x - l + x ' - 3 x + l = 0 (Trich de thi tuyen sinh sinh DH khoi DH khoi D D - 2005) 2006) CHI D A N a) Cdch 1 : B K x > - 1 , b m h phiTofng h a i v e . 2Vx + 2 + 2Vx + 1 = 4 + Vx + l (1) 4(x + 2 + 2Vx + l ) = 17 + x + 8Vx + l x>-l '4x + 8 = 17 + x o o x> - 1 Cdch 3. X = 2 : N h a n x e t x + 2 + 2Vx + 1 = (1 + yfx + lf 2(1 + V x + 1 ) - V x + 1 - 4 (1) < > = <=> x > - l b)(2)<» suy r a [Vx + 1 - 2 X = < ^x>-l 3. - x ' + 3x - 1 > 0 = - x ^ +3x-l<=> . V2x-1 2x - 1 =: (-x' + 3x - x ^ + 3x - 1 > 0 x' -6x=' + l l x ' - 8 x + 2 = 0 44 S5 TS. Vu Thg' Huu - Nguygn VTnh Can 3-V5 2 < X < 3 + V5 c:>x = l v x = 2 - V 2 . 2 ( x - l ) ' ( x ' - 4 x + 2) = 0 41. T i m m de phifofng t r i n h sau c6 h a i n g h i e m thiTc p h a n b i e t . Vx^ + m x + 2 = 2x + 1 CHI (1) (Trich de thi tuyen sink DH kiwi B - 2006) D A N (1) 1 2x + 1 > 0 o X > - - 2 f (x) = 3 x ' + (4 - m ) x - 1 = 0(*) x ' + mx + 1 = (2x + N h a n xet phu'cfng t r i n h (*) c6 he so cua x^ va he so tu do t r a i dau nen ( * ) CO h a i n g h i e m t r a i dau. TCr dieu k i e n x > - — va h a i n g h i e m ciia (*) t r a i dau, suy r a (*) p h a i c6 m o t n g h i e m thuoc nUa khoang . Dieu nay xay ra k h i m thoa m a n cac dieu k i e n sau : >0 3f S 1 4-m 1 o m>-. 2 42. >— 2 ^ 2 G i a i b a t phiTOng t r i n h 6 V x ' - 32 - 1 0 > (x - 2) (1) : x CHI D A N Bat phu'cfng t r i n h (1) l a dang cO b a n . Neu x - 2 < 0 t h i m o i x de can bac h a i c6 n g h i a deu n g h i e m diing (1). Neu x - 2 > 0 t h i (1) c6 n g h l a vdi x^ - 3x - 10 > 0. K h i do 2 ve cua (1) deu k h o n g a m , t a b i n h phiiong h a i ve t h i dau b a t phu'cfng t r i n h v a n giuf chieu. TCr l a p l u a n t r e n t a c6 the v i e t n h u sau : x-2>0 x-2<0 hoSc ( B ) (1) o (A) x' - 3 x - 1 0 > 0 - 3x - 10 > (x - 2 ) ' X (A)« < 2 x<-2 C:>X<-2 x>5 !x>2 (B) o ^ x>2 x>14 X > 14 X - - 3 x - 1 0 > x^ - 4 x + 4 Tap n g h i e m cua b a t phu'cfng t r i n h (1) l a h o p cac t a p n g h i e m cua he (A) va he ( B ) : T = (-oo; - 2 ] u [14; +oo). 43. G i a i cac bat phiJOng t r i n h : a V l 1 - X - Vx - 1 < 2 (1) i) b) Vx + 3 - V7 - X > V2x - 8 (2) Hoc va 5n luyen theo CTDT m6n Toan THPT S 45 CHI DAN ll-x>0 a) (1) c = > V l l - x < 2 + V x - l « - x - 1 > 0 o ll-x<[2 +Vx-lf 1 < x < 11 4- x< 2Vx-l '10 o < X < 4 {4 < x < l l o 1 < X < 4 22 < X < 11. b) (2)c^ Vx + 3 > V7 - X + V2x - 8 DKXD : x + 3 > 0 , 7 - x > 0 , 2 x - 8 > 0 '4 < X < 7 (2) o4 (7 - x) + (2x - 8) + 2 V 7 - x V 2 x - 8 4 < -x' + l l x - 3 0 < 0 X > 6 _Vay t a p n g h i e m cua (2) la : 4 < x < 5 u 6 < x < 7. 44. Giai b a t phirong t r i n h : V3x^ + 5x + 7 - VSx^ + 5x + 2 > 1 (1) CHIDAN o D K : 3x^ + 5x + 2 > 0 x < - 1 hoac x > — . Dat t = 3x^ + 5x + 2 thi 3 (1) t r d t h a n h : V t + 5 - >/t > 1 <=> V t + 5 > l + V t o t + 5 > l + t + 2Vt x<-l <=>Vt<2 t > 0 =>0- 2 X > t < 4 o (-2; - 1 ] u 3 TS. Vu The' - - 3 3 x ' + 5x - 2 < 0 -1 X < <=> < -2 46 ^ < X < r 2 3 — 3 Nguygn Vinh CSn Hifii - 45. G i a i bat phuong t r i n h : ^^^^^^^ + V^T^ > 4=^ (D Vx - 3 Vx ^ 3 (Trich de thi tuyen sinh DH khoi A - 2004) CHI DAN D K X D : x ^ - 16 > 0, X - 3 > 0<=>x > 4 (1) » V2(x' - 1 6 ) + x - 3 > 7 - x X >4 (A) <=> fV2(x' - 1 6 ) > 10 - 2x < X > 4 10-2x<0 x>4 10-2x>0 (B) 46. X >4 2(x'-16) >(10-2x)' He (A) CO n g h i e m : x < 5, he (B) c6 n g h i e m 10 - ^/34 < x < 5 Tap n g h i e m cua (1) l a : x > 10 - V34. G i a i bat phifofng t r i n h : a) V5x - 1 - Vx - 1 > V2x - 4 b) (1) (Trich de thi tuyen sinh DH khoi A - x + l + Vx' - 4 x + l >3V^ 2005) (2) (Trich di thi tuyen sinh DH khoi B - 2012) CHI DAN a) D K X D : 5x - 1 > 0, X - 1 > 0, 2x - 4 > 0 c:> X > 2 B i n h phiidng h a i ve (1), chuyen ve t h i c6 : x> 2 X > 2 <=> (1) « X + 2 >V2(x-l)(x-2) [(x + 2f > 2 ( x ' - 3x + 2) X > 2 x ' - lOx < 0 o 2 < X < 10. b) D K X D : x > 0 , x ^ - 4 x + l > 0 c : > 0 < x < 2-S Ta CO X = 0 l a m o t n g h i e m cua (2) X e t X > 0, chia h a i ve cho \fx t h i diroc : V x + hoSc x > 2 + Vs Vx +Jx+—- 4 >3 V X Dat t = Vx + 4 = t h i X + i - t^ - 6 t h i (2) c6 dang : Vx X t +Vt^-6>3 Vt'-6>3-t t > 2 G i a i bat phtfOng t r i n h : V x + > — t a duoc 0 < V x < —, hoSc Vx > 2 Vx 2 2 Suy r a t a p n g h i e m cua (2) l a : 0 < x < — hoSc x > 4. 4 HQC V§ 6n luy§n theo CTDT mfln Jo&n THPT El 47 1. §. HE PHUCfNG TRINH NHIEU AN 6 KIEN THLfC H $ phi^ofng t r i n h b a c n h a t h a i a n , b a a n a) H e phuong t r i n h bac n h a t h a i a n (x va y) c6 dang: (I) ajX + b i y ^ C j a 2 X + b 2 y = C2 K i hieu D = (1) (2) = aib2 - a2bi goi la d i n h thufc cila he ( I ) b2 C2 a2 C: ai b2 C2 b, Ci - = Cib2 - - aiC2 - C2bi a2Ci Quy tdc Crame. G i a i he phifcfng t r i n h bac n h a t . Neu D 0 he (I) CO nghiem duy nhat (xo; yo) xac dinh bdi cong thijfc D., y Xo = D ' _ yo = D - N e u D = 0, ^ 0 hoac Dy ;^ 0 h e . ( I ) v6 n g h i e m . - N e u D = Dx = Dy = 0 he ( I ) c6 v6 so n g h i e m l a t a p n g h i e m cua phiiong t r i n h : aix + b i y = C i hoac ciia a 2 X + b 2 y = C2. b) Gidi he phiiang trinh bac nhat hai an bdng phuang phdp do thi + b j y = c, (1) Cho he phiiong t r i n h : ( I ) - aiX [aaX + \y = C2 (2) T r e n cung m o t m a t phSng t o a do Oxy, ve cac dirorng thSng (di) c6 phiiOng t r i n h (1) v a duTcfng t h a n g ( d 2 ) c6 phtfcfng t r i n h (2). K h i do t o a do (xo; yo) ciaa giao d i e m (di) v a ( d 2 ) l a n g h i e m cua he ( I ) . N e u (di) v a ( d 2 ) giao nhau he (I) c6 n g h i e m duy n h a t . N e u d i // d2 he (I) v6 n g h i e m . Neu d i v a d 2 trCing nhau, he (I) c6 v6 so n g h i e m . Toa do m o i d i e m cua (di) (hay ( d 2 ) ) l a m o t n g h i e m . BAI TAPi 47. G i a i cac he phtfofng t r i n h sau: a) ( I ) 2x - 3y = - 4 x-2 b) ( I I ) 3x + y = 5 2 I2-X + y=7 + 5y - 3 48 S TS. Vu Thg' Hi;u - Nguyen Vinh CSn CHI D A N a) A p dung quy tac Crame cho he ( I ) . 2 -3 -4 D = = 11 ^ 0, = 3 1 5 2 -4 3 5 Dx -3 = 11, 1 = 22 = ^ . 2 2 ^ 2 —i D 11 11 N g h i e m duy n h a t cua he ( I ) l a (1; 2). Xo = D — b) Dieu k i e n xac d i n h ox ^ 2. DSt X = t h a y vao ( I I ) t a X dxsgc he phuong t r i n h bac n h a t v d i X va y. [ (ir) 3X + y - 7 - 2 X + 5y = 3 X - A p dung quy tSc Crame cho he ( I D t a difcJc y = 48. G i a i , b i e n l u a n theo t h a m so m he phiTcfng t r i n h (I) 32 17 23 17 81 32 23' y = 17 X = 6mx + (2 - m)y = 3 (m - l ) x - m y = 2 CHIDAN Ta t i n h d i n h thufc cua he ( I ) . D = 6m 2-m m -1 - m = -Gm^ - (m - 1X2 - m) = -5m^ - 3m + 2 D =0» - 5 m ^ - 3 m + 2 = 0 « m = - l hoSc m = 5 + Neu m = - 1 , D = 0, Dx = - 3 ;t 0 he ( I ) v6 n g h i e m . 2 22 + N e u m = - , D = 0, Dx = ^0 he ( I ) v6 n g h i e m . 5 5 + N e u m ^ -1, m ^ —, D = - 5 m ^ - 3 m + 2 ^ 0, he ( I ) c6 n g h i e m duy 5 n h a t (xo; yo) v d i , 3 2-m m +4 D. 2 - m Xo = D D 5m^ + 3m - 2 6m m-1 9m+ 3 D -5m^ - 3m + 2 • Hoc 6n luy§n theo CTDT mfln Jo&n THPT 0 49 49. Giai cac he phiTcrng trinh: X 2y _ 29 x + 2 y + 1 15 a) (I) 2x y 8 x + 2 y + 1 15 X + 2y + 3z = 10 b) (II) 2x + 3 y - z = l l 3x - 2y + z = 6 (1) (2) (3) CHI DAN a) DKXD: x ^ - 2 , y ^ - 1 . DM X = x + 2 Y = y + 1 thi (I) trd thanh: X + 2Y = 29 15 (D _8_ 2 X - Y = 15 3 2 Ap dung quy tSc ^Crame3 cho ^he = 2 ta tinh difdc X = —, Y = —. (I') ^ 3 , . . Giai y+1 3 ^ 5 3 x +2 5 Vay (3; 2) la nghiem cua he (I). X + 2y + 3z = 10 (1) b) (II) J2x + 3 y - z = l l (2) 3x - 2y + z = 6 (3) Nhan phiicrng trinh (1) v6i - 2 dem cong vao phiTcfng trinh (2). Lai nhan phifong trinh (1) vdi - 3 cong vao phifOng trinh (3) thi dufdc x + 2y + 3z = 10 (1') (II)i - y - 7 z = -9 (2') - 8 y - 8 z = -24 (3') Lai tiep tuc nhan phiTOng trinh (2') cua (II)i vdfi - 8 , cong vao phifong trinh (3') thi diTcfc x + 2y + 3z = 10 x = 3 (11)2 - y - 7 z = -9 y = 2. 48z - 48 z= Ghi chu: De thuc hien phep giai tren l ta c6 the lap rieng bang cac he so cua he (II) va thiTc hien nhtr sau: 'I 2 3 10^ ^1 2 3 10^ 2 3 10^ 2 3 -1 11 —> 0 -1 -7 -9 0 -1 -7 -9 6; 0 48 48j 10 -8 -8 -24; 13 -2 1 50 E3 TS. Vu The' HiAi - Nguyen Vinh C5n 2. Mpt so phxiofng t r i n h h a i a n d a n g d a c b i $ t a) He gom mot phiiang trinh bdc hai, mot phiiong trinh ax + by + c = 0 (1) (I) [ A x ' + Bxy + C y ' + D x + Ey + F = 0 (2) bdc nhdt G i a i he (I) b k n g phLrong phap the. b) He phiiong trinh dot xiing loqi I L a he phufdng t r m h co dang < l g ( x , y ) = 0 (2) Trong do f(x, y) va g(x, y) la cac bleu thufc doi xufng doi v 6 i cac bien x, y. Cdch giai: D a t a n so phu S = x + y, P = xy. Dieu k i e n can va du de he c6 n g h i e m la - 4 P > 0. rf(x,y) = 0 (1) c) He phiiong trinh doi xiing loai H: ( I I ) fly,x) = 0 (2) Cdch giai: Di/a viec g i a i he ( I I ) ve g i a i he: (F) d) Phuang trinh (III) 'f(x,y)-f(y,x) =0 f(y,x) = 0 dang cap fi(x,y) = gi(x,y) (1) f2(x,y) = g , ( x , y ) (2) T r o n g do m 6 i phuong t r i n h cua he l a m o t dSng thiJc cua cac da thufc dang cap cung bac. Cdch giai: G i a i he ( I I I ) v d i x = 0 hoSc v d i y = 0 V d i x ;t 0 dat y = k x hoSc v6x y 0 dat x = k y r o i khuf a n de doi ve g i a i phirong t r i n h m o t a n . BAI TAP 50. G i a i , b i e n l u a n theo t h a m so m he phi/cfng t r i n h 3x + 5y = 13 (1) (I) x ' + 3 y ' = m (2) CHI DAN TCf (1) X = l i z ^ y . The vao (2) t h i diroc 3 \ 13-5y + 3 y ' - m = 0 o 52y^ - 130y + 169 - 9 m = 0 (2') B i e t thufc cua (2'): A' = 65^ - 52(169 - 9m) = 4 6 8 m - 4563 4563 39 • Neu m < 468 = — , A' < 0, (2') v6 n g h i e m => (I) v6 n g h i e m . 4 HQC va 6n luy§n theo CTBT mOn Tcrin THPT S 5 1