Nối tiếp nội dung phần 1 tài liệu Phương pháp đặc biệt giải toán trung học phổ thông - Cuốn 7: Sử dụng phương pháp điều kiện cần và đủ giải toán, phần 2 giới thiệu tới người đọc các nội dung: Sử dụng phương pháp điều kiện cần và đủ giải bài toán về tính chất nghiệm, sử dụng phương pháp điều kiện cần và đủ giải bài toán về tính chất tham số. Mời các bạn cùng tham khảo.
Bai tap 15: T u n m de he sau c6 nghiem nhat: ' ' ' + l x l = y + x' +m CHUDE S L T D U N G PHU'dNG P H A P D I E U K I E N C A N V A D U | Bai tap 16: T i m m de' he c6 nghiem nhat: X" GIAI BAITOAN V E TINH CHAT NGHIEM - (2m + )x + in " + 111 = X " - x - m " +2m > M6 D A U Bai lap 17: T i m m de he c6 nghiem nhat: x^-3x 10x + > x-+2(nr-l)x -2ni + = Bai tap 18: T i m m de he sau c6 nghiem nhat: f _ x'+2x + y ^ < x- y +m=0 % IvTrong chu de sc minh hoa each str dung phuong phap dieu kien fC\n \ du giai bai toan v6 tinh chat nghiem cho phuong tiinh, bat phuong tiinh, he phuo'ng irlnh va he bii'l phuong tiinh duoc chia lhanh hai dang: Duiii^ I : Giiii bai toan Doii^ Giai bai toan ve tap nghiem 2: tinh chat cac nghiem cho phuong trinh Dang 3: Giai bai toan ve phuong trinh he qua Dan'g 4: Giai bai toan ve hai phuong trinh tuong duong i d u 2: Xac d i n h m de phuang t r i n h : BAI TOAN mx'-2(m+ GlAI B A I T O A N V E TINH C H A T C A C NGHIEM C H O PHLfaNG TRINH Gidi Dieu " Tim elicit kien ciia tluun so (i^ici si'f la ni) de pliii'cfiii^ f(.\:m) CO iii-liieni kien can: = X| + X = Iriiili: (1) • tlioci nidii liiili dial Dii'K kien 111 (*)C:>(X, +X2)'-2X,X2 = can : Gia su p h u a n g l i i n h c6 n g h i e m thoa m a n 4(111 (I) • He thiic V i e t e giua cac n g i i i e m • Bieu dien dieu k i e n K l i i o n g qua (I) • Suy dieu k i e n cho t h a m so kien di'i - irdo: t i n h ehat K , k h i ta c6: Dieu 2(in + l) m 111 + X|.X2 = K " la llurc l i i c n ihco cac buoc sau: + 1)' M ± l ) = « m = - ^ m m Dieu thay vao (1), kien du: V i - - Thuc hien phep i h u l a i 2x^ + 2x - = < = > x , = II V i DU M I N H V i (lu 1: HOA Vay, v6i m = - (6) CO hai n g h i e m x,, X2 thoa m a n 4(X| + X2) - TxiXj V i d u 3: -1 ± V3 , ihoa m a n (*) thoii m a n dieu k i e n dau bai G i a su p h u a n g t r i n h : ax^ + bx + c = (*) ;c6 hai n g h i e m x , , X2 C M R he thiic: Ciidi Dicii kien can: Gia sir phuang irinh c6 ngliicm x,, x^ Ihoa man (*} klii do: 2(m-l) 111 + b^ + a^c + ac^ = 3abc la d i e u k i e n can va d i i de phuang t r i n h c6 mot n g h i e m bang b i n h p h u a n g cua n g h i e m l a i 'Gidi m-2 X|.X, =• T h e o gia thiet ta dugc: m +1 do: S= /*\ A ( i n - l ) (=*•) - ^ Dien ta dugc: Xac d i n h m de phuang t i i n h : ( m + l)x^ - ( m - )x + m - = [ii (*) G i i i sir p h u a n g t r i n h c6 n g h i e m x,, X j t h u a man (*) k h i do: V o l yeu cau: Biivv 2: (1) =0 hai n g h i e m x, va X2 thoa m a n x, + X2 = I I ' H U O N C ; P H A P Biioc I: l)x + m + Ill + -,111-2 - m= 111 + -6 P = : ["x, = 2 Ihoa man V a y , v d i m = - thoa man dieu k i e n ddu bai — (I) c kien dir V a i m = thay vao (1), ta dugc: - 5x' + I4x - = 0 _ 9iir + 6m + > (*) nhu sau: Phuang trlnh (1) c6 nghiem phan biet lap cap so cong Vdfi dieu kien (1) c6 nghiem phan biet x,, Xj, x, thoa man : X| + X +xi ^ Bai toan tren c6 the duoc giai bang phuong phap hang so bat dinh, ( ) CO ba nghiem x,, - d, x,,, x,, + d, (d;^0) 3m • X|X2 + X X +X3X1 = -3 X1X2X3 = -3m K h i do: x^ - 3x^ - 9x + m = [x - (x„ - d)](x - x„)[x - (x„ + d)] -2 = Khi do: 15 - 2: l x - m l = x + (1) iai icii kicn can: Phuang trlnh nghiem diing Vx > - suy x = - la ghiem cua (1), tiic la: Im + 21 = » "ill = -4 Do chinh la dieu kicn can de phuang trlnh nghiem diing V x > - ^icu kicn di'i • V o i m = 0, ta co: Ixl = X + Nhan thay rang x O ^ r - , ^S) knong phai lu nghiem cua ph'J'ang trlnh do, m - khong thoa man jn = 77 • (2) nghiem diing V X G [ I , 3] ii V i m = ~ 4, ta c6: f(x) = C O nghiem x,, x, thoa man x, < < < X j Ix + 41 = X + diing vai x > - Vay, vdri m = - phuong tiinh nghiem diing Vx > - Chii y: Bai toan tren c6 the phat bieu dudfi dang: b " T i m m d6' phuong trinh Ix - ml = X (2) S/23 -m>12 Bai toan tren duoc phap bieu tuong duong dudi dang: " Tim m bat phiMng •ao (ho trinh j2\' + mx + + 15I>1 khong c6 nghiem tren Bat phuong trinh I2x^ + mx + m + 15I>1 khong c6 nghiem tien [ , 3j Bat phuong trinh I2x^ + mx + m + 151 < nghiem diing voi V x e [ l , 3] - < 2x' - 8x + < (x-2)^>0 < [x^-4x + < i du 3: » (3) Vay (1) nghiem diing V x € [ l , 3] •au: Dicii kien du: V o i m = - 8, ta c6: 2x - x + < m > - \-in < J, 3/ " K h i de giai bai toan thong thudng ta len phal bieu lai nhu kien can de (1) nghiem diing V x e [ l , 3] I2x' - 8x + 71 < « 2m + > Vay (1) nghiem diing V x e [ l , 3] m = - 22 111 [ nghiem < -8 ag(l)>0 (1) - 9< IT)- - m - i » s O A0 I2x^ + mx + m + l < 2x^-8x + > m < - g(x) = CO nghiem < < X3 < X4 Tim m de bat phuong trinh sau nghiem diing V x e [ , 3]: (1) « + 32 < [4111 g(x) = CO nghiem X3 < X4 < < Trong nghiem ciia bat phuong trinh (2) la x> - dieu J2ii + < 'g(x) = v6 nghiem + f(x) > (hoac f(x) < ) " Vay m = - la (3) nghiem diing V x e [ 1,3] tuang duong vai bat phuong trinh Vi du 2: {1[ aa ff (( 3l ))V(-)= (x+l)(4x 4x + l)>0 -3x-l0 Do chinh la dieu kien can de phuong triuli nghiem diing vdi Vx>0 )/('// kien dir V d i m = 3, (1) c6 dang: m > X>(1 , (3) nghiem dung Vx e [-2,4] V x ' + x + l = x + l x + l = x + l < = > = luon diing Vay, voi m>4 thoa man dieu kien dau bai Vay, voi m = phuong trinh nghiem diing Vx>0 Ccic/i 2: Su dung phuong phap dat an phu ciing vdri tarn thiJc bac hai f CVi/i y: V o i bai toan c6 nhieu hon mot tham so ta se thay tam quan Dat t = V(2 + x ) ( - x ) , voi x e [ - 2, 4] ta nhan dugc dieu kien ciia t la 0 - t (i) 2) chiia doan [ - , I ] " Gicii Dieu kien can: Bat phuang t r i n h n g h i e m d i i n g v a i V x [ l , 3] => n g h i e m d i i n g v a i X = 1, x = 2, tiic la ta c : I2in+17l t = ^ va t D o chfnli la dieu kien can de nghiem ciia bat phuang t i i n l i chiia [ - - < x ' - 8x + < (x-2r G i a su (1) c6 n g h i e m V t e [ ^ , > — kien du: V a i m = - 8, ta c6: (1) « kien can: - 8 Vay, voi m< - nghiem ciia bat phuang trinh chua [ - , 11 2x^-8x + < Chii y: Cung c6 the khong can su dung phuong phap dieu icien can va dii Irong bai toan tren, cu the: - < 2x' - 8x + < ix-2)^ >0 < x < x^ - x + < Vay voi m - - bat phuong trinh nghiem diing V x e [ l , 3] yi du 14: T i m dieu kien ciia m de bat phuung tiinh: Bien ddi bat phuang trinh ve dang: Ig V(2 + x ) ( - x ) ( i -t)^c:>f(t) = ( m - l ) t + < (*) Vay de nghiem cua bat phuong trinh chiia [ - , 1] dieu kien la: (1) X', ghiem diing voi moi x e ( - 2, 4) idi ^ Bien doi bat phuong trinh tuong duong voi: f(l) Do la dieu kien can de bat phuong trinh nghiem diing voi V x e ( - 2, 4) Dieu kien ccin: Gia su (1) c6 nghiem Vx > do: x = la nghiem ciia (1), Dieu kien du: Gia sir m>4, do: • ( l ) » m + = m = A p dung bat ding thiic Cosi cho ve trai, ta duoc: x/T [T-, ; 0 Dieu kien du: Vdfi m = 0, (1) c6 dang: • x>() = x o Bien doi ve phai ve dang: VP = x ' - 2x + m = (X - ) ' + m - > x = x l u n diing • Vay, voi m = phuong trinh nghiem diing voi Vx>0 Vi du 13: _ ^ = Suy ra: V(2 + x ) ( - x ) < x^ - 2x + m T i m m de bat phuong trinh sau nghiem diing V x [ 1, 3]: i x ^ m ( x i H i i < _ ( m + ) ( x ^ - x + 2) •::) (1) Vay, voi m>4 bat phuong trinh nghiem diing vdi V x e ( - 2, 4) Chu y: Co the sii dung gtln, gtnn ciia ham s6' de giai v i du tren, cu the: Gicii Dieu kien can: Bat phuong trinh nghiem diing V x e [ l , 3] => nghiem diing voi X = x = 2, tiic la ta c6: I2m+17l0 i d u 3: = -4 vc// moi xeD " ) C h u n g ta d i x e m Cho p h u o n g t r i n h va bat p h u o n g t r i n h : > < in < Vx-l+2nWx-2 (1) + Vx-i-2nWx-2 = (2) Ix^ + 3x + 2l3x = k7rox=-!^, R G Z Bai tap 2: T i m gia tri ciia m de hai phuang trlnh tuang duang 2cosx.cos2x = + cos2x + cos3x Do ( I ) & (2) tuang duang trudc het can x = (la mot nghiem cua ho X = ~ ) cung la nghiem ciia (2), tire la: 4cos^x - cos3x = acosx + (4 - a)( + cos2x) Bai tap 3: T i m gia tri ciia m de hai phuang tnnh tuang duang m.cosO + Iml.cosO + cosO = m + Iml = m < cos3x = 4cos(37t + x) Do chinh la dieu kien can ciia a va b mcos^x + (1 - m ) s i n ( ^ + x) = Dicu kien dit • Bai tap 4: V a i m = 0, ta dugc: (2) ! - 2sin^3x = sin3x = » x =k7r«x= sin3x = ainx + (4 - 2a)sin^x ~.k&Z Bai tap 5: Xac dinh a, b de hai phuang trlnh sau tuang duang: asin2x -2-^3 = a A/3 Vay m = thoa man dieu kien dau bai • ' tuong duang vai bat phuang trlnh Vx^ +4x + + x + 1>0 - 2.sin^3x + 2msin3x.sinx = (sin3x - msinx)sin3x = Bai tap 7: (De 10): T i m m de hai phuang trlnh sau tuang duang: sin 3x = sin X - sin 2cos2x.cosx = + cos2x + cos3x x- ni sin x = 4cos^x - cos3x = mcosx + (4 - m)( + cos2x) Bsii tap 8: (De 40): Tun m de hai phuang trlnh sau tuang duang: X = 3x - kn - 4sinx Bai tap 6: T i m m de phuang trlnh: lx-ml-lx+11 = m.co.s2x - m.co.s4x + cos6x = i cos6x - - m(co.s4x - cos2x) = o COSX sin2x + cos2x + b V2 + = 42 sinx + 2cosx(co'sx + b) Vdri m0 x-l>0 3-Vx-l « l < x < >0 Bien doi (2) ve dang: V6-X = 3- V x ^ < » V6-X + >/x^ =3 ( - x ) + ( x - l) + V ( - x ) ( x - l ) = o II v i D U M I N H H O A V(6-x)(x-l) X = Vi (111 1: o T i m x de phuong trinh sau nghiem diing vdi moi a: log^,2^2^N/x + - l ) = l0g^2^2^2(2-V5^) Gidi (1) b (6 - x)(x - 1) = « x' - 7x + 10 = o = 2 x =5 Su dung dieu kien can va dii de thuc hien Dieu kien can: Phuang trinh nghiem dung vo'i Vm > 0, truac het nghiem Dieu kien can: Gia sir (1) nghiem dung voi mgi a => diing voi a = V a i a = 0, ta duoc: dung vai m = "x = ( l ) o l o g ( V ^ - l ) = log2(2- ^/^) 0, bai vcfi in = mau thuin 10? 103 • V o i X = 5, phucfiig t r i n h (1) c dang: BAI T O A N l o g j l = log2^„l (luon diing) HE N G H I E M D U N G V O l V a y , X = n g h i e m d i i n g phucnig t r i n h da cho v d i m o i m > III G l A TRI XAC D I N H CUA T H A M SO BAITAPDfiNCJHI I F H i r O N C ; P H A P Biii tap 1: Cho p h u o n g t r i n h 2log^^2_^2^4-V7 + 2x) = log^^2^2^2^4-3x) a G i i i i phuong t r i n h v d i a = b T i m cac gia tii cua x ngliiem diing phuong tiinli da cho voi moi m > Bai tap 2: Cho p h u o n g t r i n h l o g ( a V + 5a^x + V - x ) = log O - V x - l ) a Cho he chiia hai t h a m so a, b, v d i yeu cau: " Tim a de lie c6 nghiem vai moi h lliiigc D,, " ta ihuc hien theo cac budc: Bum-1: Dat d i u k i e n de cac bieu thuc ciia he CO ngliTa Bum-2: Dieii kien can: G i a sir he n g h i e m d i i n g v d i V b e D ^ , suy no n g h i e m d i i n g v d i b„eD,„ G i a i p h u o n g t r i n h v d i a = • • b T u n cac gia t i i ciia x ngliiem diing phuong tiiiih da cho vol m o i m > Q Biioc 3: G i a i he v d i b = b„ => g i a t r i ciia a„ Dicht kien dir Thuc hien phep k i e m tra v d i a = a„ II V i D U M I N H H O A Vidul: T i m a de he phuong t r i n h sau CO n g h i e m v d i m o i b: (x^ + i r ' + ( b ' + l ) y = a + bxy + x"y = Giai Dicht kien can: D o he c d n g h i e m v d i V b , nen phai c d n g h i e m k h i b = K h i d d he CO dang: (x- + ir' = a + x^y = fa = O JL^ =0 a + x"y = : a=0 a = 1' V a y dieu k i e n can la a = hoac a = Dieu • kien can ^ V d i a = he c d dang: (b2 + i)y = i (1) bxy + x'^y = (2) He tren k h o n g the c d n g h i e m v d i m o i b, bdi k h i h^O (i)^y = o N g h i e m k h o n g thoa m a n (2), d o vay he v n g h i e m 105 • N h i l n xet rang he tren c h i c6 n g b i e m k h i I - 2b > b < ^ Vdfi a = he c dang: V a y a = k h o n g thoa m a n • V d i a = - , k h i he c6 dang: bxy + x ' y = he l u o n c6 n g h i e m x = y = v d i m o i b V a y , v o i a = he c6 n g h i e m voi m o i b V i d i i 2: Xac d i n h cac gia t r i ciia a cho he sau day c6 n g h i e m v o i moi b ^ - o g X + log2y = I N h a n xet rang he tren l u o n nhan x = & y = l a m n g h i e m V a y v o i a = - he c n g h i e m v d i m o i b Vx^-2b^-l-v'(a-l)by =x-l (I) ax + by - = Vi dii 4: V a i ' a > 0, cho he p h u o n g t r i n h : x + y = b^-b+ l Gicii DieII J Vx" »- -1 = (I) « ax - = T i m a de he c6 n g h i e m v a i m o i b e [ , l ] ] ^ X =l [ax - = [a = Gidi B i e n d o i t u a n g d u a n g he ve dang: V a y a = la dieu k i e n can de he c6 n g h i e m v o i m o i b Dieu a-^+ay=2 kien can: He c6 n g h i e m v o i m o i b => c6 n g h i e m v o i b = 0, k h i do: b^-b+l kien di'i: V d i a = , he ( I ) c6 dang: fx-i>0 Vx^-2b^-l = X -1 X + by - = X = b^ + I « \2 X « + by - = aUa>'=r X = b^ + X + by - = i fx - b " + It nha't mot n g h i e m la • ^ ~ [b-'+by = ' ' K h i a\* la n g h i e m d u a n g ciia p h u a n g t r i n h : t ^ - l t +a^'-^^' = kien can: H e c6 n g h i e m \6\, 1] • Dii'u p h u a n g t r i n h ( ) c6 hai n g h i e m d u a n g V b e [ , 1] iy = - b V a y , he p h u o n g t r i n h c6 n g h i e m v i m o i b k h i a = Vi d u 3: a\a-^=a*^'-''^' => p h u a n g t r i n h ( ) c6 hai n g h i e m d u a n g v d i b = va b T i m a de' he sau c6 n g h i e m v d i m o i b: 2'''"'^-^'+b(a + l)log.^y ^ a ^ • V a i b = , theo a) ta c6 d i e u k i e n l i i < a < • V a i b = ^ , d i e u k i e n la p h u a n g t r i n h (I) — 16 ( a - l ) l o g X + log5y = Gidi Dieu kien can: He c6 n g h i e m v a i V b => c6 n g h i e m v i b = 0, k h i do: (I)« 1^- l = a" => a = ± 4t ( a - l ) l o g X + log2y = D o c h i n h la d i e u k i e n ciin de he n g h i e m v a i V b Dieu • kien di'i iog2 y = 106 2hl"y3''+2b = l [iog^ y = =0 i 4a4 >0 A>0 V a i a = , k h i he c6 dang: 2''''*'%2blog^y = l + a"* hai n g h i e m d u a n g S>0 b'''"'^-^'= l - b I log2 y = P>0 o i > a>0 c^O0, V b e [ , 1] Do chinh la 6'i6u kien can de he nghiem voi Vbe [0, 1] Xet ham so y = b' - b + (1 + log J 6) Dieu kien dir Gia sir < a < —\j= , k h i xet (1) ta c6: 32V2 Voi Vbe [0,1] • Mien xac dinh D = [0, 1] • Dao ham: y' = b - 1, A he c6 nghiem Vay, voi < a < — — • c^y_ >0c;>^+log l6>0«a0 A>0 s > , V b e [ , 1] o f>0 P>0 Giiii he vdi a = b = b Tim a de he sau co nghiem vdi moi b Hai tap 2: , V b e [ , 1] Xac dinh cac gia tri ciia a cho he sau cd nghiem vdi moi b: la(x~+y") + x + y = b a'''-''-'>0 c:> a''"-''-' < — , V b e [ , 1] 16 He C O nghiem vd; V b e [0 ] C O nghiem vdi b = => (theo a) < a < — , la dieu kien can ciia a 108 a y-x =b (3) T A I I.II;:U TIIAm K i i A o Tnin Van Hao ( chu bien ) - Cluiyen de luyen thi vao Dai hoc Dai so - Niia xuat ban Giao due -2001 Nguydn Van Mau - Phuong piicip giai phuor.g innli va bat plurong trinhNha xuat ban Giao due - 2001 Phan Huy Khai - Phirong phap dieu kien cttn va du de bien luan he c6 tham so - Nha xuat ban Giao due - 2001 NguySn Dire DOng & Nguyen Van VTnh - 15 phuong phap chuyen de tarn tlu'rc bac va cac ling dung dae sflc - Nha xuat ban Tie - 2000 TiSn Phirong - Le Hong Diic Dai so so cap Nha xuat ban Ha No! - 2003 Phuorng phap dac biet giai toan T H P T sif Dmci i>iiLrOi\fii I M I A P » H U K i f \\iV v A » i j G i A i TOA^ Chiu tntch nhiem xuat ban : Gidm doc N C U Y f i N K H A C O A N H Bic'ii tup noi clung: L E BiCH NGOC Tiinli bay hia: L E SICH NGOC Che ban: L E HUtJ T R I L I CUOI Nhdm CiiMon luon san long gmi dap mgi tlidc mdc cua cac em hoc sinh Vii dpcgj'a ve noi dung cua cuon hii lieu Mgi clii tiet xiii lien lie true tiep tcfi: Ths L e Hong Due So nha 20 - Ngo 86 - Duong T o Ngoc Van - Tay Ho - H a Noi Dien tlioai: 04 7196671 ... - ( m + l ) ' - ( m + l ) + 2m = » m = I 68 _g(X) = x- - (3in - )x - 3m - - Viet lai phuong trinh ve dang: (X - l ) [ x - - (3m - l ) x - 3m - 2] = "x = l g(x) = x - - ( m - l ) x - m - = (2)' ... x e [ - 4, 6] 90 ^ 7x-+4xy + 2y < ! 2lll-l 2in • + 5x "-4 xy+ 2y- > tap 17: T i m m de he sau c6 nghiem: 15x ^-1 lxy + y - = - X V x - +2mx > X + + 6m + < 3^ (x - 3>' - m - 2)( x - 3^ - m - 5)