DANG 1: TUONG GIAO, KHOANG CACH, GOC Sur twang giao: Cho 2 do thi ciia ham sd: y = f(x), y = g(x) - Phuong trinh hoanh do giao diem: f(x) = g(x) <o f(x) - g(x) = 0 la mot phuong trinh dai sd, tuy theo sd nghiem ma cd quan he tuong giao: vd nghiem: khdng cd diem chung, 1 nghiem (don): cat nhau, 1 nghiem kep: tiep xuc, 2 nghiem: 2 giao diem,.... - Nghiem phuong trinh bac 3: ax3 + bx2 + ex + d = 0, a * 0 Neu cd nghiem x = x 0 thi ta phan tich thanh tich sd: (x - x0 ) (Ax 2 + Bx + C) = 0 Neu dat ham sd f(x) = ax3 + bx2 + ex + d thi dieu kienxd 1 nghiem: dd thi khong cd cue tri hoac y C o • ycr > 0, cd 2 nghiem: yen ycT = 0, cd 3 nghiem phan biet: yco ycr < 0. Chuy: - Phuong trinh bac 3 luon ludn cd nghiem. y C D . y C T <0 - Phuong trinh bac 3 cd 3 nghiem duong khi: < x C D , x C T > 0 a.f(0) < 0 - Hai diem tren 2 nhanh do thi y = . ta thudng lay hoanh dd k - a x - k va k + b vdi a, b > 0 Goc: - Gdc giua 2 vecto: cos(u , v ) = +y y - Gdc giua 2 dubng thang: cosa = | cos( n, n')' - Neu he sd gdc cua 2 dudng thang la k, k' thi: tan(D,D') Khoang each: V A 2 + B 2 VA' 2 + B k-k ' 1 + kk' - Giira 2 diem: AB = ^(x B - x A ) 2 +(y B - y A ) 2 - Tu M0(x0, y0) den (A): Ax + By + C = 0: d = ^ +By° + °l N/A 2 + B 2 - Dd thi ham bac 3: y = f(x) cat true hoanh tai 3 diem A, B, C theo thu rued khoang each AB = BC tuc la 3 nghiem xu x2 , x3 lap cap so cdng thi diem udn thudc true hoanh. - Phuong trinh trung phuong ax4 + bx2 + c = 0, a * 0 cd 4 nghiem phan biet lap cap sd cdng khi 0 < t x < t 2 , t 2 = 9tt . -BDHSG DSGT12/1- 149 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
V j du 1: Chung minh rang vdi moi gia tri cua m, dudng thang y = x - m + 1 • -X2 + 3x - 1 . . . , A cat y = tai hai diem phan biet. x - l Giai Hoanh do giao diem cua dudng thang va dudng cong la nghiem cua —x2 + 3x — 1 2 phuong trinh = x - m + 1 co 2x - (m + 3)x + m = 0 , x * 1. x - l 2 2 V i x = 1 khong phai la nghiem va A = (m + 3) - 8m = m - 2m + 9 > 0 vdi moi m, nen phuong trinh luon cd hai nghiem phan biet suy ra dpcm. V i du 2: Tim m de do thi ham so sau cat true hoanh tai 3 diem phan biet: y = x3 + (2m + l)x 2 + (3m + 2)x + m + 2. Giai Cho y = 0 co x3 + (2m + l)x 2 + (3m + 2)x + m + 2 = 0. co (x + l)(x2 + 2mx + m + 2) = 0. CO x = - 1 hoac f(x) = x2 + 2mx + m + 2 = 0 (1) Dd thi cua ham sd da cho cat true hoanh tai ba diem phan biet khi va chi khi phuong trinh (1) cd hai nghiem phan biet khac - 1. fA' = 0 [m 2 -m-2> 0 , , _ < co < co m < - 1 hoac m > 2, m * 3. [f(-l)* 0 [-m + 3*0 V i du 3: Tim m de dd thi (C„): y = x3 - 3mx + m + 1 cat true hoanh tai 3 diem phan biet. Giai D = R. Ta cd y' = 3x2 - 3m, y' = 0 <=> x2 = m. Dieu kien (C m ) cat true hoanh tai 3 diem phan biet la: m > 0 vayco-ycT < 0 <=> f(-Vm ).f(Vm ) < 0. co (m + 1 - 2m. Vm )(m + 1 + 2m Vm ) < 0 co (m + l) 2 - 4m3 < 0. co -4m3 + m2 + 2m + 1< 0 co (m - l)(4m2 + 3m + 1) > 0 co m > 1. (vi A = 9 - 16 < 0 nen 4m2 + 3m + 1 > 0, Vm). V i du 4: Tim m de dd thi (C m ): y = xJ - 3x2 + 3(1 - m)x + 1 + 3m cat true hoanh dung 1 diem. Giai D = R. Ta cd y' = 3x2 - 6x + 3(1 - m), A' = 9m. - Neu m < 0 thi y' > 0, Vx nen (C m ) cat true hoanh dung 1 diem: thoa man. - Neu m > 0 thi y' = 0 cd 3 nghiem phan biet xi , x 2 va S = 2, P = 1 - m. DK (C m ) cat Ox dung 1 diem la yco-ycT > 0. Ta cd y = - (x - l)y' + 2(-mx + 1 - m) => yi = 2(-mxj + 1 + m) nen dieu 3 kien la 4(-mx! + 1 + m)(-mx2 + 1 + m) > 0. co m2 X!X2 - m(l + m)x1 x2 + (1 + m) 2 > 0 150 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
<=> m2 (l - m) - 2m(l + m) + ( l + m) 2 >0om 3 <lom<l . Vay gia tri can tim la m < 1. Vi du 5: Vd i cac gia tri nao cua m, duong thang y = m + 3x cat do thi y = x 4 - 2x2 + 3x - 3 tai bon diem phan biet. Giai Phuong trinh hoanh do giao diem cua dudng thang va dudng cong x4 — 2x2 + 3x — 3 — m + 3x <=> x4 — 2x2 — m — 3 = 0 (1) Dat X = x2 , X > 0, ta duoc: X 2 - 2X - m - 3 = 0 (2) Duong thang cat dudng cong da cho tai bon diem phan biet khi phuong trinh (1) cd bdn nghiem phan biet, dieu nay tuong duong vdi phuong trinh (2) cd hai nghiem duong phan biet. A'> 0 [m + 4> 0 P > 0 co <Um-3> 0 co -4<m<-3 . S >0 2> 0 Vi du 6: Cho ham sd y = x 4 - (3m + 2)x2 + 3m cd dd thi la (C m ), m la tham sd. Tim m de dudng thang y = - 1 cat dd thi (C m ) tai 4 diem phan biet deu cd hoanh dd nhd hon 2. Giai Phuong trinh hoanh dd giao diem cua (C m ) va dudng thang: x 4 -(3 m + 2)x2 + 3m = -l . Dat t = x2 , t > 0, phuong trinh Ud thanh: t 2 - (3m + 2)t + 3m+ 1 =0. co t = 1 hoac t = 3m + 1. Yeu cau cua bai toan tuong duong: f 0< 3m + l< 4 1 < co — < m < 1, m * 0. [3m + l * l 3 Vi du 7: Tim cac gia tri cua tham sd m de dudng thang y = -2x + m cat do thi ham sd y x2 + x - 1 tai hai diem phan biet A, B sao cho trung diem cua doan thang AB thudc true tung. Giai Phuong trinh hoanh dd giao diem: x2 + x - 1 -2x + m co 3x2 + (1 - m)x - 1 = 0 (x * 0). Vi a, c trai dau nen phuong trinh cd hai nghiem phan biet xi , x 2 khac 0 vdi moi m. Hoanh dd trung diem I cua AB: x m - 1 2 6 Dieu kien IeOycox i = 0com = l . Vi du 8: Tim cac gia trj cua m de dudng thang (d m ) di qua diem A(-2; 2) va co 2x- l he so gdc m cat dd thi cua ham sd: y = x + 1 -BDHSG DSGT12/1- 151 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
co m < 0 hoac m > 12. a) Tai hai diem phan biet? b) Tai hai diem thuoc hai nhanh cua do thi? Giai Phuong trinh cua dudng thang (d m ) la: y = m(x + 2) + 2 co y = mx + 2m + 2. Hoanh do giao diem cua dudng thang (d m ) va dudng cong la nghiem ciia 2x - 1 phuong trinh: mx + 2m + 2 = x + 1 <o (mx + 2m + 2)(x + 1) = 2x - 1, x * - 1 co mx2 + 3mx + 2m + 3 = 0, x * - 1 (1) a) Duong thang (d m ) cat dudng cong da cho tai hai diem phan biet khi va chi khi phuong trinh (1) cd hai nghiem phan biet khac -1 . \a*0 [m * 0 [A>0 ,g(-1)* 0 ^ |m 2 -12m> 0 b) Hai nhanh cua dudng cong da cho nam ve hai ben cua dudng tiem can dung x = - 1 cua dd thi. Duong thang (d m ) cat duong cong da cho tai hai diem thudc hai nhanh cua nd khi va chi khi phuong trinh (1) cd hai nghiem Xi, x 2 va x i < - 1 < x2 . Dat x = t - 1 thi X, < - 1 < x2 => ti < 0 < t 2 . Phuong trinh trd thanh: m(t - l) 2 + 3m(t - 1) + 2m + 3 = 0 co mt 2 + mt + 3 = 0 (2). Dieu kien phuong trinh (2) cd hai nghiem trai dau coP<0com<0 . Cach khac: (1) co m(x2 + 3x + 2) = -3. Xet m = 0: loai. 3 Xet m * 0 thi x 2 + 3x + 2 = nen dieu kien m 3 xi<-l<x 2 c o > f(-l ) = 0 co m < 0. m Vi du 9: Tim theo m, xet su tuong giao cua dd thi: y = dudng thang y = x + m. Giai Phuong trinh hoanh dd giao diem cua 2 dd thi: -3/2 - 2x2 -2x- 3 2x2 -2x-3 x-3 Taed 9 + 3(1 = x + m co x2 + (1 - m ) x - 3(1 - m) = 0 (x * 3) (*) - m) - 3(1 - m) * 0 vdi x * 3. A = (1 - m) 2 + 12(1 - m) = (m - l)(m - 13), do do: Neu m > 13 hoac m < 1 Neu m = 1 hoac m = 13 A > 0 nen cd 2 giao diem. A = 0 nen cd 1 diem chung la tiep didm. 152 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Neu l<m<13=>A< 0 nen khong co diem chung. Vj du 10: Tinh gdc giua cac tiem can cua dd thi ham sd y x2 + x- 2 x + 1 Giai x 2 +x- 2 2 Tac6y = = x — , D = R\ {-1} . x + 1 x + 1 Vi lim y =+oo, lim y =-co nen dd thi cd TCD: x =-1. x-»(-ir X-K-D* 2 Ta cd lim (y - x) = lim = 0 nen TCX: y = x. x->±* x->ta x + 1 Vi TCD vudng gdc vdi Ox nen gdc giua TCX va TCX la 45° Vi du 11: Tim m de tiem can xien cua dd thi (C) mx2 + 2(1 - 2m)x + 3m -1 , ,. . „0 y = hop voi Ox goc 30 x - 3 Giai „ , mx2 +2(l-2m)x + 3m-l , _ 5 Ta co y = = mx + 2 - m + x - 3 x - 3 5 V i lim (y - (mx + 2 - m)) = lim = 0 nen TCX la y = mx + 2 - m x->+x x-»±°c x- 3 vdi m * 0, cd he sd gdc k = m. R TCX hop vdi Ox gdc 30° co m = ±tan30° = ± — 3 , , x 2 + 2mx — 1 Vi du 12: Tim m de duong thang y = 2m cat dd thi ham sd y = x - l tai hai diem phan biet M va N sao cho OM _L ON. Giai Duong thang y = 2m cat dd thi ham sd tai hai diem M va N khi phuong trinh hoanh dd cd 2 nghiem phan biet xi, x2 khac 1: x2+2mX ~1 = 2mcox2 -l + 2m = 0.Dodd:m<-,m*0. x - l 2 , (2x + 2m)(x-l)-(x2 +2mx-l) ,, , 2m Taco y' = ^ — =>y(xi) = (x-l ) x; T^-I , •» ^r , S^-K, 2m 2m , 4m Dieu kien OM _L ON <=> = - 1 <=> = - 1 Xj x2 X1X 2 —1 + >/5 <=> 4m2 + 2m - 1 = 0 co m = — = (chon). 4 ' ,,, . .„ , , i C( . x2 - x + m2 + 3m - 2 , Vi_du_13: Cho ham so y = t(x) = 1 im cac gia tn cua x + 2 tham sd m de dd thi cua ham sd cat true hoanh tai 2 diem phan biet la A va B sao cho dp dai cua doan thang AB bang 1. -BDHSG DSGT12/1- 153 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Giai Phuong trinh hoanh do giao diem vdi true hoanh. x2 - x - m2 + 3m - 2 = 0, x * -2 Ta cd: A = 4m2 - 12m + 9 = (2m - 3) 2 Dd thi (1) cat Ox tai 2 diem phan biet khac 2 fA> 0 12m-3* 0 f 3 <=> { co < co nu ! -1 ; - ; 4 lg(-2)* 0 -m 2 +3m-4* 0 [ 2 Ta cd AB = | x : — x2 1 = V A = V4m2 -12m + 9 Nen AB = 1 <=> 4m2 - 12m + 9 = lcom 2 - 3 m + 2 = 0 <=> m = 1 hoac m = 2 (chon). V i du 14: Tim cac gia tri cua tham sd m de duong thang y = - x + m cat dd x2 - 1 thi ham sd y = tai hai diem phan biet A, B sao cho AB = 4. x Giai Phuong trinh hoanh dd giao diem X 2 - l . '.„. 2 -x + m co2x -mx - 1 = 0, x* 0 (1). x Vi a. c trai dau nen (1) cd hai nghiem phan biet Xi, x 2 khac 0 vdi moi m. Goi A(xi; yi), B(x2 ; y 2 ) nen: AB2 = (xt - x,) 2 +'(y1 - y 2 ) 2 = 2(x: - x2 ) 2 ' Ap dung djnh li Viet ddi vdi (1) ta duoc: 2 A B2 = 2[( X l + x2 ) 2 - 4 X l x 2 ] = — + 4. 2 Do do AB = 4 co — + 4 = 16 co m = ±2 S . 2 V i du 15: Tim cac gia tri cua m sao cho dd thi cua ham sd y = x4 - (m + 1 )x2 + m cat true hoanh tai bdn diem, tao thanh ba doan thang cd dd dai bang nhau. Giai Hoanh dd giao diem cua duong cong va true hoanh: x 4 - (m + l)x 2 + m = 0 co x2 = 1 hoac x2 = m. Dieu kien m > 0 va m * 1. Khi dd. phuong trinh cd 4 nghiem x = -1 , x = 1, x = - Vm , x = Vm o o 1 1—' i 1 1 1—1—4- 1 Duong cong cit true hoanh tai 4 diem tao thanh ba doan thang bing nhau khi: Vin = 3 hoac Vm = - co m = 9 hoac m = i (chon). 3 y 154 s -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Vi du 16: Tim m de ducmg thang y = m cat do thi y = x4 - 2x2 + 2 tai 4 diem phan biet cd hoanh do tao thanh mot cap so cong. Giai Phuong trinh hoanh do giao diem: x4 - 2x2 + 2 = m. Phuong trinh cd 4 nghiem phan biet tao thanh mot cap so cong, ttic la doan AB = BC = CD. Bai toan dua ve tim m sao cho phuong trinh cd 4 nghiem phan biet la ±x0 va ±3x0 (x0 > 0), tire la: 0 5 [ x 4 -2x 2 + 2 = m |81x 4 -18x 2 + 2 = m m v . _ 41 Vay m = — 41 25 25 Cach khac: Dat t = x , t > 0 rdi dua ve dieu kien t 2 = 9t|. Vi du 17: Chung minh cac dudng thang d: y = m - x ludn cat dd thi (C): x2 - 3x x - l tai 2 diem M , N va cat 2 tiem can cua (C) tai P, Q ddng thoi hai doan MN, PQ cd ciing trung diem. Giai Phuong trinh hoanh do giao diem d va (C): X ~3x =ra-xo2x2 -(m + 4)x + m = 0,x#l. x - l Ta cd x = 1 khdng la nghiem va A = m2 + 16 > 0, Vm nen d ludn cat (C) tai 2 diem phan biet M , N. x 2 - 3 x „ 2 Ta cd y = nen TCD: x = 1, TCX: y = x - 2. Do x - l x- l do x P = 1, hoanh dd giao diem Q cua d vdi TCX: m - x = x - 2 XQ m + 2 .Dodd Xp+X * m + 4 dpcm. 2 2 2 2 Vi du 18: Chung minh tich cac khoang each tir diem M bat ki thudc dd thi (C) ddn 2 tiem can la mdt hang sd. a)y = 2x + l x - 3 b)y Giai 4x + 5x - 4 x + 2 a) Dd thi y = Vdi M(x; 2x + l x - 3 2x + l cd TCD A: x = 3, TCN A':y = 2. ) e (C), tich khoang each den 2 tiem can. -BDHSG DSGT12/1- 155 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
d(M; A).d(M, A') = | x-3 1 2x + l „ 7 x-3 |. 2 = x-31. x - 3 x - 3 = 7 khdng ddi 4 x 2 +5x- 4 b) y = : = 4x - 3 + x + 2 x+ 2 TCX A': y = 4x - 3 <=> 4x - y - 3 = 0. 2 nen TCD A: x = -2, Vdi M(x; 4x - 3 +- x + 2 d(M; A).d(M, A') = | x + 2 ) e (C), khoang each den 2 tiem can: 2 2 Vl6 + l.|x + 2| 7l7 : khdng ddi. V i du 19: Chung minh dd thi ham sd: y x2 + (2m + l)x + m2 + m + 4 ludn 2(x + m) ludn cd cue dai, cue tieu va khoang each giua cue dai, cue tieu khdng ddi. D = R\ {-m} . Taed y' Giai x2 + 2mx + m2 2(x + m) 2 y' = 0 o x2 + 2mx + m2 - 4 = 0, x * -m. 3 Ta cd A' = 4 > 0, Vm va g(-m) = -4*0 , Vm nen hai cue tri A(-m-2; --) , B(-m + 2; -). 2 Do dd khoang each AB = Vl6 + 16 = 4 V2 : khdng ddi. V i du 20: Tim m de dd thi ham sd: y = x3 - 3x2 + mx + 1 cd cue dai, cue tieu va hai diem do each deu dudng thang d: y = -2m. Giai D = R. Ta cd y' = 3x2 - 6x + m. Dieu kien cd CD va CT la A' = 9 - 3m > 0 <=> m < 3. Ta cd y = (—x - — )y' + 2(- m - l)x + - m + 1 nen dudng thang qua 2 3 3 3 3 diem CD, CT la d': y = 2( - m - l)x + - m + 1. 3 3 Dieu kien CD, CT each deu d: y = -2x la d' hoac song song vdi d hoac d di qua trung diem 1(1; m - 1) cua doan ndi CD, CT. 2(-m - l) = -2, -m + 1 * 0 hoac m - 1 =-2 . 1 . 3 3 <=> m = 0 hoac m = - 1 (chon). 4x- 3 V i du 21: Tim diem M thudc dd thi (C): y - ——— cd tdng cac khoang each X O den 2 tiem can be nhat. 156 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Giai 4x — 3 Do thi y = co TCD A: x = 3, TCN A': y = 4. x 3 4x - 3 Goi M(x; ) e (C), ta co d(M; A) + d(M; A') x - 3 + x - 3 4x-3 x - 3 - 4 = x- 3 + x - 3 >2V9 =6 Dau "=" xay ra khi va chi khi | x - 3 | = , 9 n l <=> (x - 3) 2 = 9, do dd cd 2diemM(6; 7) va M'(0; 1). Vi du 22: Tim diem M thudc dd thi (C): y = 2 true be nhat. x - 3 x 2 - 3 x-2 cd tdng khoang each den Giai x 2 - 3 Goi M(x; ) e (C) thi tdng khoang each deh 2 true d = |x|+ x - 2 x 2 - 3 x - 2 3 3 3 x * 2. Xet diem A(0; —) e (C) thi d = —, do dd min d < — nen chi xet 2 2 2 cac diem cd hoanh dd x < — 2 . x 2 - 3 Khi do x - 2 > 0 nen d = x x 2 - 3 x-2 2x2 -8 x + 7 (x-2)2 NeuO<x< - thid = f(x) = x+- — -,f'(x V 2 x- 2 v ; f'(x) = 0 o x = 2 - —.Lap BBT thi min d = f(0) = - 2 2 Neil-- <x<Othid = g(x) = -x+ 2Lzi g'(x)= ~1 ,<0 2 x- 2 5 (x -2) 2 Do do g nghich bien tren --; 0 bg(x)>g(0 ) 3 3 So sanh thi min d = - tai M = A(0; -) . 2 ' 2 Vi du 23: Tim diem M thudc dd thi (C): y = x - l x + 1 cd tdng khoang each den 2 true be nhat. x-l Giai Goi M(x; x + 1 ) e (C), tdng khoang each den 2 true la d = | x | + x - l x + 1 -BDHSG DSGT12/1- 157 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
x * -1 . Xet diem A(0; 1) e (C) thi d = 1 nen min d < 1, do dd chi xet cac x-l diem cd: x < 1, x + 1 d = x + =x - 1 + x + 1 Dau "=" khi x + 1 < 1 nenO<x < 1, khi do: 2 ....... x + 1 = -2 + (x + 1)+ >-2 + 2^2 x + 1 x+ 1 <=> (x + l)2 = 2 c=> x = -1 + V2 Vay co 2 diem M(- l - V2 ; 1 + V2 ) , (-1 + V2 ; 1 - V2 ). 2x + 3 V i du 24: Tim hai diem tren 2 nhanh do thi (C): y = — cd khoang each x - l be nhat. Giai „ , t 2x + 3 Q ^ 5 Ham so y = — = 2 + . x * 1. x - l x - l Gpi A( 1 + a; 2 + - ) , B(l - b; 2 - - ) la 2 didm tren 1 nhanh dd thi vdi a b a, b > 0. Ta cd BA2 = (a + b) 2 + 5 5 a + b Ap dung bat dang thuc Cdsi thi BA2 > 4ab + — > 2V8O = &S ab 20 Dau "=" khi a = b va 4ab = — c=> a = b = Vo" ab Vay A(l + Vo"; 2 + V5) , B(l-</5 ; 2 + Vo"). • x2 — x — 1 V i du 15: Tim hai diem tren 2 nhanh dd thi (C): y = — cd khoang x - 2 each be nhat. Giai T T , i X" - x - 1 , 1 Ham so y = : — = x + 1 + . x * 2. x x - 2 Gpi A(2 + a; 3 + a +- ) , B(2 - b; 3 - b - - 1 -) la 2 diem thudc 2 nhanh vdi a b a, b > 0. Ta co: BA2 = (a + b)2 + fa + b + - + ^j =(a + b)2 = 8 + 4 2ab + — | > 8 + 4.2V2 I ab ' 158 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Dau = khi a = b va 2ab = — <=> a = b ab 1 1 VayA(2 + ^ ; 1 V2 1 72+-^)vaB( 2 V2 Vi du 26: Tim diem M thuoc (P): y = f(x) = -3x2 + 8x - 9 va N thuoc (P'): y = g(x) = x2 + 8x + 13 sao cho M N be nhat. Giai Ta cd khoang each M N be nhat khi 2 tiep tuyen tai M va N song song vdi nhau va chung vudng gdc vdi doan MN. Goi M(x; f(x), N(xi, g(x,)) thi f'(x) = g'(x,) o -6x + 8 = 2xi + 8 <=> x, = -3x. Do do MN 2 = 4(36x4 - 192x3 + 392x2 - 352x = h(x). Ta cd h'(x) = 64(9x3 - 36x2 + 49x - 22) = 64(x - l) 2 (9x2 - 27x + 22) h'(x) = 0 <=> x = 1. Lap BBT thi minh(x) = h(l) = 5. Khi do M(l ; 4), N(-3; -2), kiem tra M N vudng gdc vdi 2 tiep tuyen tai M, N: dung. Vay M(l ; 4), N(-3; -2). DANG 2: TIEP TUYEN, TIEP XUC Tiep tuyen: Cho dd thi (C): y = f(x) - Tiep tuyen tai diem M(xo;yo): y - yo = f '( x o) ( x _ x o)- Phuong trinh nay co 3 yeu td x0 , yo va he sd gdc: f '(x) = k = tan(0x,t) - Tiep tuyen di qua A(x A , y A ): y - y A = k(x - x A ) <=> y = g(x) Tim he sd gdc k bang each giai he phuong trinh cho tiep diem: f(x) = g(x) f'(x) = g'( x ) Cach khac: lap phuong trinh tiep tuyen tdng quat tai xo vdi an xo rdi cho qua A thi tinh duoc xo. Tiep xiic: Cho 2 dd thi y = f(x) va y = g(x). Dieu kien tiep xuc la he , , ff(x ) = g(x) . .. . phuong trinh: \ , , co nghiem B [f'(x ) = g'(x) Chii y: Vdi hai dudng thang d: y = ax + b, d': y = a'x + b' thi cd: d • d' khi a = a', b = b'; d // d' khi a = a', b * b'; d 1 d' khi a. a' = - 1 jra -BDHSG DSGT12/1- 159 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
V i du 1: Viet phucmg trinh tiep tuyen cua do thi tai diem udn: a) y = x3 + 3x2 - 4 b) y = -x 4 + 2x2 - 2 Giai a) D = R. Ta cd y' = 3x2 + 6x, y" = 6x + 6 y" = 0 co x = - 1 nen dd thi cd diem udn I(-l ; -2). Phuong trinh tiep tuyen tai M0 ( x0 ; f(x0 )) y = f '(x0)(x - x0 ) + f(x0 ) = -3(x + 1) - 2 nen y = -3x - 5. b) D = R. Ta cd y' = -4x3 + 4x, y'' = -12x2 + 4,y" = 0 o x - nen dd thi cd hai diem udn: I 13 va J 13) Suy ra phuong trinh tiep tuyen ciia dd thi tai hai diem udn dd la: 8 7 , y = j= x — va y 8 7 r^X 3x/3 3 3V3 3 V i du 2: Cho dd thi (C): f(x) = 2x3 + 3x2 + 1 va (P): g(x) = 2x2 + 1. a) Viet phuong trinh cac tiep tuyen ciia (C) va (P) tai mdi giao diem ciia chiing. b) Xac dinh cac khoang nao tren do (C) nam phia tren hoac phia dudi (P). Giai a) PT hoanh dp giao die'm: 2x3 + 3x2 + 1 = 2x2 + leo x2 (2x + 1) = 0 co x = 0 1 • 1 3 hoac x = — Suy ra hai giao diem: A(0; 1) va B( — ; —) 2 2 2 Ta cd f *(x) = 6x2 + 6x; g'(x) = 4x f '(0) = 0; g'(0) = 0 nen dudng thang y = 1 la tiep tuyen chung cua (C) va (P) tai diem A. 3 3 3 — nen tiep tuyen ciia (C) tai diem Bla y = -— x + — -2 nen tiep tuyen cua (P) tai diem B la y = -2x + — b) Lap hieu sd: f(x) - g(x) = 2x + x = x (2x + 1). Do dd, tren khoang (-00; — ) 2 thi (C) nam phia dudi (P) cdn tren cac khoang (— ; 0) va (0; +00) thi (C) 2 nam phia tren (P). 1 x2 V i du 3: Cho 2 dd thi: y = f(x) = -==, y = g(x) = -== xv2 v2 a) Tim giao diem cua 2 dd thi. b) Lap phuong trinh cac tiep tuyen cua hai dd thi tai giao diem. Tinh goc hop bdi 2 tiep tuyen do. 160 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Giai 1 x2 a) Phucmg trinh hoanh do giao diem — = = -== , x * 0. xV2 V2 3 1 cox = 1 co x = 1. Suy ra giao diem M(l ; -j=). v2 b) Ta cd f '(x) = ,_ => f '(x) = —]= nen tiep tuyen cua dd thi ham sd f V2.x2 V2 tai M lay = —= (x-2). V2 Ta cd g'(x) = x V2 og'(l) = 42 nen tiep tuyen cua dd thi ham g tai M la y = V2 x — \ = . V2 Ta cd f '(l).g'(l) = - 1 nen 2 tiep tuyen do vudng gdc nhau, suy ra gdc 2 tiep tuyen la 90° Vi du 4: Lap phucmg trinh tiep tuyen vdi dd thi: a) y = 2x3 - 6x2 + 3 va cd he sd gdc be nhat. b) y = -x 3 + 3x2 va cd he sd gdc ldn nhat. Giai Ta cd he sd gdc cua tiep tuyen la dao ham tai do. a) y' = 6x2 - 12x = -6 + 6(x - l) 2 > -6, dau "=" khi x 0 = 1 nen max y' = - 6 , do do tiep tuyen tai A(l ; -1) la y = -6x + 5. b) y' = -3x2 + 6x = 3 - 3(x - l) 2 < 3, diu '"-' khi x 0 = 1 nen min y' = 3, do dd tiep tuyen tai D(l ; 2) lay = 3x - 1. Vi du 5: Lap phuong trinh tiep tuyen tai x = 1 vdi dd thi ham sd y = f(x) thda man f 2 ( l +2x) = x - f 3 ( l -x) . Giai Lay dao ham 2 ve, ta cd: 4f(l + 2x) .f'( l + 2x)= l + Sfa-x ) f'(l-x) . The x = 0: 4f(l) .f'(1) = 1 + 3f 2 (x) . f'(1) (*) The x = 0 vao f (1 + 2x) = x - f( l - x) => f(l ) = -f (l) . =0 + f(l)) = 0 =0 f(l) = 0 hoac f( l) = -1 . Vdi f(l) = 0 thi (*): 0 = 1 (loai) Vdi f(l) = 1 thi (*): -4f'(D = 1 + 3f '(D =>f'(1) = Y Vay phuong trinh tiep tuyen y = —- (x - 1) Vidu 6: Lap phuong trinh tiep tuyen dd thi (C): y = x3 - 5x2 + 2: a) Song song vdi dudng thang y = -3x - 7. b) Vuong gdc vdi dudng thang y = - x - 4. -BDHSG DSGT12/1- 161 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Giai Taxd y' = 3x2 - lOx a) Tiep tuyen song song voi duong thang y = -3x + 1 nen: y' = -3 co 3x2 - 10x = -3 co 3x2 - 10x + 3 =0 ox0= -hoac x0= 3. 3 1 .u- 40 .x x , , 67 . , . Voi x 0 = — thi y 0 = — : tiep tuyen y = -3x + — (chon) 3 27 2 V Vdi x0 = 3 thi y0 = -16: tiep tuyen y = -3x - 7 (loai). b) Tiep tuyen vuong goc vdi dudng thang y = — x - 4 nen: 7 y' = -7 co 3x2 -lOx = -7 co 3x2 -lOx + 7 = 0 co x0 =1 hoac x 0 = - 3 Vdi x0 = 1 thi y0 = -2: tiep tuyen y = -7x + 5 Voi x0 = — thi y0 = : tiep tuyen y = -7x + 3 27 27 Vi du 7: Viet phuong trinh cac tiep tuyen cua do thi ham so: x2 -x + l . „ , , A .„ a) y = va vuong goc voi tiem can xien. x - l b) y = -x3 + 3x2 - 2 va song song vdi d: y = -9x + 25. Giai x2 - x + 1 1 a) Ta cd y = = x + , x *1 nen TCX: y = x. x - l x - l Tiep tuyen cua do thi vuong gdc vdi tiem can xien cd he so gdc lc.yW..101__i_r ..1,„IO_i_-2 co (x - l)2 = — co x0 = 3 + -= hoac x0 = 1 •= 2 V2 V2 Tu do lap duoc 2 tiep tuyen: y = -x + 2±2\/2 b) Goi (d) la tiep tuyen phai xac dinh. He so gdc cua (d) la k = -9. Hoanh do tiep diem M cua (d) vdi (C) la nghiem cua phuong trinh y'(x) = -3x2 + 6x = -9 co x2 - 2x - 3 = 0 co x0 = -1 hoac x0 = 3. Vdi x0 = -1, phuong trinh tiep tuyen: y = -9x - 7 (chon) Vdi x0 = 3, phuong trinh tiep tuyen: y = -9x + 25 (loai) V i du 8: Viet phuong trinh tiep tuyen ciia do thi (C): y = - X + x + 1 v ^ & x + 1 X" + X + 1 uong trinn uep luyen eua uu uu ^j . y = quaM(-l;0). Giai Duong thang d qua M(-l; 0) co he so gdc k: y = k(x + 1) Dieu kien d tiip xuc vdi (C) la he sau phai cd nghiem: 162 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
x2 + X + 1 f(x) = g(x) f'(x) = g'(x) ° < x + 1 x2 +2x (x + 1)2 = k(x + l) k c x^x+ l x2 +2x , , , 3 Suy ra = co x0 = 1, do do k = - x+1 x+1 4 3 Vay PT tiep tuyen qua M la y = — (x + 1). 4 Vi du 9: Lap phuong trinh tiep tuyen vdi do thi (C): y = x3 - 5x2 + 2 a) Tai A(0; 2) b) Di qua A(0; 2). Giai Ta cd: y' = 3x2 - lOx. Phuong trinh tiep tuyen tai diem M(XQ; y0 ) y = f '(x0) .(x-x 0 ) + yo a) Tai A(0; 2) nen x0 = 0, y0 = 2, f '(x0) = 0, ta cd tiep tuyen: y = 0(x - 0) + 2 = 2. b) Ta cd y 0 = x^ - 5x2 + 2, f '(x0 ) = 3 x2 -10xo nen phuong trinh tiep tuyen tai M(x0 ; y0 ) bit ki la: y = (3 x^ - 10xo)(x - x0 ) + (x3 , - 5 x2 + 2). Cho tiep tuyen qua A(0; 2) la 2 = (3 x2 - 10xo)(0 - x0 ) + (x3 , - 5 x2 + 2) o2xj-5xo 2 =0ox2 (2x0 - 5) = 0 co x0 = 0 hoac x0 = - 2 Vdi x0 = 0 thi cd tiep tuyen y = 2. 5 25 Vdi x0 = — thi cd tiep tuyen y = x + 2. 2 4 Vi du 10: Lap phuong trinh tiep tuyen cua dd thi (C): y = j2-x biet tiep tuyen cat Oy tai B(3; 0). Giai Vdi 2 — x > 0 <=> x < 2 thi y' 2^2- x Phuong trinh tiep tuyen tai M(x0 ; y0 ) Cho tiep tuyen qua B(3;0): 0 = (3-x 0 ) + J2-x 0 2^2-:- x co 3 - x0 - 2(2 - x0 ) - 0 co x0 = 1 (chpn) Vay tiep tuyen can tim: y = -- (x - 3). Vi du 11: Lap phuong trinh tiep tuyen chung cua 2 dd thi: y = f(x) = -x 2 - 2x + 1, y = g(x) = x2 - 2x + 3. -BDHSG DSGT12/1- 163 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Giai Ta co f '(x) = -2x - 2, g'(x) = 2x - 2. Duong thang y = ax + b la tiep tuyen chung ciia 2 do thi tai M(c; f(c)) va N(d; g(d)) a = f(c ) = g'(d) ac + b = f (c) ad + b = g(d) |-2c- 2 = 2d- 2 {-c 2 - 2c + 1 + 2c(c + 1) = d2 - 2d + 3 - 2d(d -1) f'(c) = g'(d) f(c)-cf'(c ) = f(d)-d.g, (d) CO CO [c + d = 0 c 2 + d 2 = 2 CO c = l , d = - l c =-l , d = l Tir do cd 2 tiep tuyen chung: y = 2 va y = -4x + 2. Vi du 12: Chung minh tiep tuyen tai A(-l ; 0) cua do thi (C): y = -x" + 2x" + x cung la tiep tuyen cua do thi nay tai mot diem B khac A nua. Giai Ta cd y' = -4x 3 + 4x + 1. Vdi x0 = -1 , y 0 = 0 thi f '(x0 ) = 1 nen tiep tuyen tai A(-l ; 0) lay = x + 1. Dat y = f(x) = -x 4 + 2x2 + X ; y = g(x) = x + 1. De tiep tuyen tai A cung la tiep tuyen tai B khac A thi he sau cd nghiem ff(x) = g(x) Xo*-l : -x4 + 2x2 + x = x + 1 f'(x)=g'(x ) -Ax6 + 4x + l = l CO - x 4 +2x 2 1 = 0 [fx 2 -l) 2 = 0 CO < v -4x J + 4x = 0 c o x = +l . 4x(x2 -1) = 0 Chon nghiem x 0 = 1 * - 1 nen B( 1; 2)=> dpcm. Chu y: Day la tiep tuyen di qua 2 tiep diem. V i du 13: Cd bao nhieu tiep tuyen cua dd thi (C): a ) y = x +3x + 3 ^ A ^_j . ^ x + 1 b)y x 2 + l di qua B(-l ; 7). Giai a) Dudng thang d di qua A(-l ; 0), he sd gdc k cd phuong trinh y = k(x + 1). De dudng thing d la tidp tuydn ciia (C), dieu kien can va du la he phuong ' x 2 +3x + 3 trinh sau cd nghiem: • x + 1 x2 +2x = k(x + l) <=> < (x + 1)2 -3 ,2 . X gx co < 3 (x + 1)2 164 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
3 3 Vay tir A ve dugc mot tiep tuyen den (C): y = — x + — b) Phuong trinh dudng thang qua diem B he so gdc k cd dang d: y = k(x + 1) + 7. (d) la tiep tuyen cua (C) khi va chi khi he sau cd nghiem: x + —= k(x + l) + 7 x 1 ^ = k x = o < k + 7 k 2 + 18k+ 45 = 0 k = -3,x0 = - ° 2 -15,x„ = — Vay tir B ve duoc 2 tiep tuyen den (C): y = -3x + 4, y = -15x - 8. Vi du 14: Chung minh khdng cd tiep tuyen ciia dd thi (C): y = di qua giao diem cua 2 tiem can. 1 x2 +2x + 2 x + 1 Ta cd y = x + 1 Giai x * - 1 nen cd TCD: x 1, TCX: y = x + 1. x + 1 giao diem 2 tiem can I(-l ; 0) Phuong trinh dudng thang (d) qua I vdi he sd gdc k la y = k(x + 1). Gia su d la tiep tuyen cua (C) thi he sau cd nghiem. k(x + l) = x + l + - ^ x + 1 k = l - 1 1 (x + 1) 1- 1 (x + 1) 1 (x+ir ) = x + l + - 1 x + 1 (X*-1): = 0 : vd li x + 1 x+1 ' x+ 1 Vay khdng mdt tiep tuyen nao cua (C) di qua I . Vi du 15: Cho ham sd y = -x 3 + 3x2 - 2 cd dd thi (C). a) Tim tat ca nhirng diem tren dudng thang y = 2 ma tir do cd the ke duoc 3 tiep tuyen den (C). b) Tren (C) cd nhung cap diem ma tai dd 2 tiep tuyen ciing cd he sd goc p, chung minh trung diem cua cac doan thang ndi tirng cap diem dd la diem cd dinh. Giai a) Lay M(a; 2) thudc dudng thang y = 2. Duong thang qua M cd he sd gdc k cd dang y = k(x - a) + 2. Goi A(x0 ; y0 ) la tiep diem cua duong thang va (C) thi Xo la nghiem cua he: |-x 3 +3x 2 - 2 = k(x 0 - a ) + 2 (1) { k = -3x2 + 6x0 (2) ( X o - 2) [2x2 - (3a - l)x 0 + 2] = 0 (3) -BDHSG DSGT12/1- 165 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Dieu kien co 3 tiep tuyen voi (C) ve tu M la (3) co 3 nghiem phan biet tuong ung 3 nghiem phan biet ciia (2): 5 19a' -6a-15> 0 l8-2(3a-l ) + 2* 0 a < 1 hay a > • a *2 b) Tidp tuyen vbi (C) co he so goc p, hoanh do tiep diem la nghiem phuong trinh: y' = p <=> 3x2 + 6x = p o 3x2 - 6x + p = 0 (4) A' = 9 - 3p > 0 <=> p < 3. Vdi p < 3 thi (C) cd 2 tiep tuyen song song vdi he sd gdc p. Goi xi , X2 la nghiem ciia (4), vdi 2 tiep diem M], M2 thi trung diem M1M2 cd hoanh do: X| 1 => y, = 0 Vay trung diem MiM 2 la diem cd dinh 1(1; 0). V i du 16: Chung minh tren (C): y = x3 - 2x2 + 2x + 9 khdng cd hai diem nao ma hai tiep tuyen tai hai diem dd vudng gdc nhau. Giai Ta co: f '(x) = 3x2 - 4x + 2 > 0, Vx e R. Goi Xi, x 2 la hoanh dd hai diem bat ki tren (C) thi he sd gdc hai tiep tuyen vdi (C) tai hai diem tren la f '(xi) va f '(X2). Ta cd: f '(xi), f'( x 2 ) > 0 nen f '(xi). f '(x2 ) * -1 . Vay hai tiep tuyen nay khdng the vudng gdc vdi nhau. V i du 17: Chung minh tiep tuyen tai diem M bat ki thudc do thi (C): — Vx-4x 2 y cat true tung Oy tai mdt diem A each deu gdc O va tiep diem M. Giai Vdi dieu kien x-4x 2 >0<=>0<x < - thi: y' l-8 x trinh tiep tuyen tai M(x0 ; y0 ) la: y = 4 4vx-4x 2 l-8x„ Phuong 4Vx P -4x2 2 -4x2 Cho x = 0 thi y Ta cd: l V X o - 4 X o nen tiep tuyen cat Oy tai A(0; 4Vx 0 -4x 0 2 A M = 1 (Xo-0)2 + Xo 2 16(x0 -4x2 ) V i du 18: Viet phuong trinh tiep tuyen cua (C): y = x3 - 3x2 + 1 tai didm udn I . Chung minh rang tren khoang (-00; 1) dudng cong (C) nim phia dudi = AO=> dpcm 166 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
tiep tuyen do va tren khoang (1; +00) dudng cong (C) nam phia tren tiep tuyen do. Giai Ta cd: f '(x) = 3x2 - 6x; f "(x) = 6x - 6. f "(x) = 0 co x = 1 nen diem udn 1(1; -1) Phuong trinh tiep tuy en cua (C) tai didm I la y = -3x + 2. Dat g(x) = -3x + 2, ta cd: f(x) - g(x) = x3 - 3X2 + 1 - (-3x + 2) = x3 - 3x2 + 3x - 1 = (x - l) 3 Vi f(x) - g(x) < 0 tren khoang (-00; 1) nen (C) nam phia dudi tiep tuyen va f(x) - g(x) > 0 tren khoang (1; +co) nen (C) nam phia tren tiep tuyen dd. x + 1 Vi du 19: Xac dinh m de dudng thang d: y = 2x + m cat (C): y = tai x - l hai diem phan biet A, B sao cho cac tiep tuyen cua (C) tai A va B song song vdi nhau. Giai -2 .1 Ta cd: D = R \ {1} , y' = - Phuong trinh hoanh dd giao diem cua ( x -l) 2 dudng thang (d) va dd thi (C): x + 1 = 2x + m co 2x2 - (3 - m)x - m - 1 = 0, x * 1. V i phuong trinh x - l khdng cd nghiem x = 1 va A = (m + l) 2 + 16 > 0, Vm nen PT ludn cd hai nghiem phan biet, do do (d) cat (C) tai hai diem phan biEt A, B. Tiep tuyen tai A va B song song khi: -2 -2 ( x A -l) 2 (x B -l) ! co x A - 1 = -(x B - 1) co x A + x B = 2 co = 2 com = -l . Vi du 20: Cho ham so y = f(x) = x + - cd dd thi (C). x Tidp tuyen cua (C) tai diem M(x0 ; f(x0 )) cat tiem can dung va tiem can xien tai hai diem A va B. Chung minh rang M la trung diem cua doan thang AB va tam giac OAB cd dien tich khdng phu thudc vao vi tri cua diem M tren (C). Giai Dd thi (C) cd TCD: x = 0 va TCX: y = x f ( x ) = i . Phuong trinh tiep tuyen cua (C) tai diem M(x0 ; f(x0 )) la: x ML -nnusc DSGT12/1- 167 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Diem A: x = 0 => y A DiemB: y =0 v XcJ (- X o) + Xo +— = — J Xo Xo i ^ 1 x 2 -n (x-x 0 ) + x 0 + — = x«—-+ — - 0 x„ X o Xo _ rr\ r 0 + 2X„ X. + XR O X B = 2x 0 . Ta co: XM = x0 = 2 2 V i ba didm A, M , B thang hang nen M la trung diem cua doan AB. Dien tich tam giac OAB la: Q 2 _ 1 I || I 1 S-gWIxB l — 2x = 2 , vdi moi x 0 * 0. V i du21 : Cho ham so y = x - 1 - k(x - 1) (1) a) Tim k de do thi ham so (1) tiep xuc vdi true hoanh. b) Tim k de tiep tuyen tai giao diem A vdi true tung, tao vdi cac true toa do mot tam giac cd dien tich bang 4. Giai a) Dd thi (1) tiep xuc vdi true hoanh ung vdi k sao cho: y = 0 x 3 -l-k(x-l ) = 0 <=> < co y' = o 3x 2 - k = o k = 3 , 3 b) Cho x = 0=>y = k - l^ > A(0; k - 1) Ta cd y' = 3x2 - k, y'(0) = -k. Phuong trinh cua tiep tuyen tai A: y - (k - 1) = -k(x — 0) <=> y = —kx + k — 1. Tiep tuyen nay cat true Ox tai B k - 1 Theo gia thiet: S0 A B = 4 co — |k -l | k k - 1 ; 0 , vdi k * 0. 4cok 2 -2k+ l = 8k . Xet k < 0, k > 0 giai ra duoc 4 gia tri: k=5±2V 3 va k = -3 + 272 V i du 22: Chung minh hai dd thi sau tiep xuc nhau: 13 a) y = f(x) = x3 + — x + l va y = x2 + 3x - 1 4 b) y = f(x) = —^x2 + x + - va y = (x)=> / xz - x + 1 Giai a) Hoanh do tiep diem cua hai dudng cong la nghiem cua he phuong trinh: 168 -BDHSG DSGTUllDownload Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
ff(x) = g(x) If'(x) = g'(x) 3 . 13 .. . 1 ..2 x + — x+l=x+3x+ l 4 2 13 3x2 + — = 2x + 3 4 co s x 3 - x 2 + - = 0 4 1 <=> x = — , 1 2 3x2 -2 x = 0 4 .'15 Vay hai duong cong tiep xuc vdi nhau tai diem M( —; —) . 2 4 b) Hai do thi tiep xuc nhau khi va chi khi he sau cd nghiem: i 2 i n •X +x + — = VX 4 — X + 1 = - 2 2V^ x + 1 2x- l (D (2) fx" - x + 1 The (1) vao (2): (x - l)(x2 - 5x - 6) = 0 co x = ±1 hoac x = 6. Chon x = 1 la nghiem cua he => dpcm. x2 3 Vi du 23: Chung minh rang cac do thi ciia hai ham sd: f(x) = — + —x va ?(x) 3x x + 2 chung tai tiep diem tiep xuc vdi nhau. Viet phuong trinh tiep tuyen chung cua Giai Hoanh dd tiep diem cua hai dudng cong la nghiem cua he phuong trinh: x2 3 3x — + — x = 2 2 x + 2 V 3 — + —x v 2 2 ( co < x2 3 3x —+—x = 2 2 x + 2 3x x + 2 3 x + — = • 6 Ta cd (1) co x = 0 x + 3 co 2 (x + 2) 2 x = 0 x2 +5x = 0 (D (2) x = 0 x = -5 2 x + 2 Suy ra he phuong trinh cd mdt nghiem duy nhat x = 0. Vay hai dudng 3 cong tiep xuc vdi nhau tai gdc toa do O: y'(0) = — Phuong trinh tiep tuyen chung cua hai dudng cong tai diem chung lay = — x. Vi du 24: Chiing minh 3 dd thi cua ba ham sd: f(x) = -x2 + 3x + 6, g(x) = x3 - x2 + 4 va h(x) = x2 + 7x + 8 tiep xuc vdi nhau tai diem A(-l; 2). Giai -BDHSG DSGT12/1- 169 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Ta co diem A(-l ; 2) la didm chung cua ba dudng cong da cho. Ngoai ra, ta cd: f '(x) = -2x + 3 ; g'(x) = 3x2 - 2x ; h'(x) = 2x + 7 nenf'(-l) = g'(-l) = h'(-l) = 5. Vay ba dudng cong cd tiep tuyen chung tai diem A => dpcm. V i du 25: Xac dinh parabol (P): y = 2x2 + bx + c tiep xuc vdi hypebol (H): y = — tai diem M( —; 2). x 2 Giai Ta cd: M e (P) nen b + 2c = 3. (P): y = 2x2 + bx + c =o y' = 4x + b ^> y'(^) = b + 2 (H):y=- =>y' = -- oy'(^) = -4. x x 2 Parabol tidp xuc vdi hypebol tai diem M khi va chi khi: b + 2c = 3 , „ 9 co b = -6; c = — b + 2 = -4 2 Vay (P): y = 2x2 - 6x + | Vi du 26: Tim m de phuong trinh I x3 - ^-x2 - m | = x2 - 2x + 1 cd 3 nghiem phan biet. Giai PT o | x3 - -- x2 - m | = (x - l) 2 o x3 - — x2 - m = ±(x - l) 2 2 2 3 1 co x3 x2 + 2x - 1 = m hoac x3 + — x2 - 2x + 1 = m. 2 ' 2 Xet f(x) = x3 x2 + 2x - 1, g(x) = x3 + -x 2 - 2x - 1 thi 2 dd thi cua 2 2 chung tiep xuc nhau tai A(l ; ^) . Ve 2 dd thi tren cung 1 he true toa do, suy ra dieu kien dudng thang y = m cat 5 15 1 3 diem lam e 27 2 2 DANG 3: YETJT6 CO DINH, B6 l XUNG - QUY~ TlCH Diem dac biet cua ho do thj: (C m ): y = f(x,m) - Diem A(x A , y A ) e (C m ) <=> YA = f(x A , m). Neu ta coi f(x A , m) - y A = 0 la phuong trinh theo an m thi so gia tri tham sd m la so do thi di qua diem A. - Diem cd dinh cua ho la diem ma moi do thi deu di qua: M0 ( x 0 , yo) 6 (C m ), Vm co y 0 = f(xo, m), Vm 170 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
- Diem ma ho khong di qua la diem ma khdng cd do thi nao cua ho di qua vdi moi tham sd: Mo(x0 , y 0 ) g (C m ), Vm o y 0 * f(x0 ; m) Vm • Phirong phap: Nhdm theo tham sd va ap dung cac menh de sau: Am + B = 0,Vmo A = 0,B = 0 A m2 + Bm + C = 0, Vm co A = 0, B = 0, C = 0 Am + B * 0, Vm <=> A = 0, B * 0 A m2 + Bm + C * 0, Vm <=> A = 0, B = 0, C * 0 hoac A * 0, A = B2 - 4AC < 0 Yeu to doi xung: Su dung dinh nghTa ve tam ddi xung, true ddi xung. - Ham sd chan: Vx e D => - x e D va f(-x ) = f(x) Dd thi ham sd chan ddi xung nhau qua true tung. - Ham sd le: Vx e D =o - x e D va f(-x ) = - f(x) Dd thi ham sd le ddi xung nhau qua gdc O. - Cdng thtic chuyen he true bang phep tinh tien OI. (Oxy) -» (IXY) vdi I(x0, y0): fX = * + X° r [y = Y + y 0 - Dieu kien (C) nhan I(x0 , yo) la tam ddi xting. f( x n - x) + f(x n + x) ., ^ , , , yo = — - Vx 0 - x, x 0 + x e D, hoac chuyen true bang phep tinh tien den gdc I ndi tren la ham sd le. - Dieu kien (C) nhan d: x = a lam true ddi xung; f(a - x) = f(a + x), Va - x, a + x e D, hoac chuyen true bang phep tinh tien den S(a,0) la ham so chan. Chuy: - Ham bac hai cd true ddi xiing la dudng thang di qua cue tri va vudng goc voi Ox: x = 2a - Ham bac ba cd tam ddi xiing la diem udn. - Ham huu ti 1/1, 2/1 cd tam ddi xung la giao diem 2 tiem can. Cd 2 true doi xting khdng song song vdi Oy la 2 phan giac ciia gdc hop bdi 2 tiem can. - Diem A ddi xiing B qua I khi I la trung diem doan AB. - Diem A ddi xung B qua dudng thang d khi d la trung true cua doan AB. - Ta cd the du doan yeu td ddi xiing qua tap xac dinh, bang bidn thien, dang do thi, nhdm sd hang.... Phuong phap tim quy tich diem M : Tim toa do x, y cua M , khu tham sd giua x va y. Gidi han: Chuyen dieu kien ndu cd cua tham sd ve dieu kien ciia x (hay y). Dac biet: Ndu M(x,y) e (V) thi chi can tim x rdi nit tham sd dd the, khu tham sd. ,-BDHSG DSGT12/1- 171 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
V i du 1: Tim diem cd dinh cua cac dd thi a) y = x3 + mx2 - m b)y = Giai x2 -2(m + 2)x + 6m + 5 x - 2 Goi M(x0 ; y0 ) la toa dd diem ma dd thi di qua Vm. a) Ta cd y 0 = xf, + mx2 , - x 0 - m , Vm co (x 2 - l)m - y 0 + x2 - x 0 = 0, Vm. x 2 = l Jo x? -x „ •o = - 1 . v o = 0 Vay dd thi cd hai didm cd dinh M(l ; 0) va M'(-l ; 0). 1 N ^ , x 2 -2( m + 2)x n+ 6m + 5 w b) Ta cd y 0 = -2 . Vm <=>y0 = 2(3-x o )m + x 2 -4x 0 + 5 . Vm 3-x o = 0 , x o -2* 0 y0 • 4x„ + 5 |x„= 3 l y D = 2 x 0 - 2 Vay M(3; 2) la diem cd dinh cua cac do thi. V i du 2: Tim didm cd dinh cua cac: ax2 +(l-a) x + l- a a) TCX cua dd thi y = x + 1 a*0. b) Duong thang qua CD, CT cua dd thi: y = mx3 - 3mx2 + (2m + l)x + 3 - m. Giai a a) Ta cd y = ax + 1 - 2a + x + 1 x * - 1 vdi a * 0. V i lim (y - (ax+1 - 2a)) = lim = 0 nen TCX: y = ax + 1 - 2a. X->±co X + 1 Ta cd y = a(x - 2) + 1 nen cac TCX ludn di qua diem cd dinh M(2; 1) vdi moi a*0. b) y' = 3mx2 - 6mx + 2m+l,A>0com< 0 hoac m > 1. Ta cd y — y'+ 2 2tn x + lg_J] l n dn dudng thang qua CD, CT la 3 3 3 2-2 m 10 - m m 2x-1 0 , . , ; y = x + = (2x +1) + —-— . tu do suy ra duong thang 3 3 3 3 di qua CD, CT qua didm cd dinh A(--; 3). -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
V j du 3: Chung minh rang vdi moi gia tri cua m, duong thang y = mx - 2m - 4 ludn di qua mdt didm cd dinh cua dudng cong (C): y = x3 - 3x2 + 2x - 2. Giai Ta cd y = mx - 2m - 4 = m(x - 2) - 4. Vdi moi gia tri cua m, dudng thing da cho ludn di qua diem cd dinh A(2; -4). V i toa dp cua didm A thoa man phuong trinh y = x3 - 3x2 + 3x - 2 nen A thudc (C): dpcm. Vi du 4: Chung minh cac dd thi y = (1 + m)x3 + 3(1 + m)x2 - 4mx - m ludn di qua 3 diem cd dinh thang hang. Giai Gpi M(x0 ; y0 ) la didm cd dinh cua cac dd thi: V0 = (1 + m) xf + 3(1 + m) x0 2 - 4mx0 m, Vm. o y Vm. 0 = (xf +3x f - 4x 0 -l)m + xf + 3 xf ^ jx f +3x o 2 -4x o - l = 0 (1) | y 0 =xf +3x o 2 (2) Ta chung minh (1) cd 3 nghiem phan bidt. Xet ham sd f(x) = x3 + 3x2 - 4x - 1 thi f lidn tuc tren R, ta cd f(-6) = -85 < 0, f(-l) = 5 > 0, f(0) = - 1 < 0, f(2) = 11 > 0 nen (1) co 3 nghidm phan bidt thudc 3 khoang (-6 -1), (-1; 0), (0; 2). Tii (1) => xf + 3 xf = 4x0 + 1 nen (2) => y 0 = 4x0 + 1. Vay 3 diem cd dinh thang hang tren dudng thang y = 4x + 1. Vi du 5: Tim nhung diem ma moi do thi khdng di qua: a) y = x3 + (m + l)x 2 + (m2 - 3m)x + (5 + 2m - m2 ) , . mx + 9 b ) y = — x - 9 Giai a) Gpi diem M(x; y) la diem ma moi dd thi khdng di qua <=> phuong trinh y = f(x) vd nghiem Vm (1). Ta cd y = x3 + (m + l)x 2 + (m2 - 3m)x + (5 + 2m - m2 ) o m2 (x - 1) + m ( x 2 - 3x + 2) + (x3 + x2 + 5 - y) = 0. x - l = 0 x2 -3 x + 2 = 0 x 3 + x 2 +5-y* 0 [x-l*0 ' [A = (x2 - 3x + 2) 2 - 4(x - l)(x3 + x2 + 5 - y) < 0 Vay nhung diem can tim nam tren dudng thang cd phuong trinh x = 1. bd diem A(l ; 7) va cac diem M(x; y) sao cho: (x - l)[(x - 1) (x - 2) 2 - 4(x3 + x2 + 5 - y)] < 0. nen (1) o hoac -BDHSG DSGT12/1- 173 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
b) Goi M(x0 ; y0 ) la cac diem ma moi do thi khong di qua: " x 0 =9 x 0 * 9 , mxo +9*y o ( x o -9),Vm mx0 +9 w y 0 * — — . Vm co co *o- 9 x o = 9 Vay tap hop cac diem can tim la 2 dudng thang: x = 9 va x = 0, bd diem A(0;-1). V i du 6: Tim cac diem tren dudng thang d: x = 1 ma cac dd thi y = x3 - 3mx2 + (2m2 - l)x + m2 - 5m + 1 khdng di qua. Giai Goi M(l ; y) e d la didm can tim: y * 1 - 3m + 2m2 - 1 + m2 - 5m + 1, Vm. co 3m2 - 8m + (1 - y) * 0 <=> A = 16 - 3(1 -y) < 0 co y < - — Vay cac diem can tim la M(l ; y) vdi y < - 13 V i du 7: Chung minh cac dd thi ham sd y = —— ^ X + m . m * 0 ludn tiep x - m xuc vdi 1 dudng cd dinh tai mdt didm cd dinh. Giai x,. k , (m-l) x +m Goi M(x0 ; y0 ) la diem co dinh: y0 = . Vm * 0 x Q - m o (m - l)x0 + m = yo (x0 - m), x0 * m, Vm * 0. co (x0 + 1 + y D )m - x 0 (l + y0 ) = 0, x0 * m, Vm * 0 co x0 = 0, y0 = -1 . Ta cd y' = -m , x * m => y'(0) = -1 . (x-m) 2 Vay cac dd thi ludn ludn tiep xuc nhau tai didm cd dinh M(0; -1). cd tiep tuyen chung y = - x - 1. V i du 8: Chung minh rang cac do thi ham sd sau cd true ddi xung: a) y = -x 2 -x- 3 2 c) y = x4 + 4x3 + 4x2 b) y = -3x4 + x2 - 1 Giai a) y' = x - 1. y' = 0 co x = 1. Dd thi cd diem cue tri la 1(1; — ) , Chuyen he true bang phep tinh tidn theo vecto OI: < x=X+l v=Y-7 ' 2 174 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
The' vao ham s6: Y - - = - (X + l) 2 - (X + 1) - 3 co Y = -X 2 2 2 2 Vi Y - F(X) = —X2 la ham so chan nen do thi doi xung nhau qua true tung IY cd phuong trinh: x = 1. b) Vi y = f(x) = -3x4 + x2 - 1 la ham so chan nen do thi doi xung nhau qua true tung Oy. c) y' = 4x3 + 12x2 + 8x = 4x(x2 + 3x + 2) y' = 0 co x = -2 hoac x = -1 hoac x = 0. Xet diem I(-l; 1). Chuyen he true bang phep tinh tien theo vecto — fx=X-l OI: [ y= Y+i The vao ham sd: Y +1 = (X - l)4 +4(X - l)3 +4(X - l)2 co Y = X4 - 2X2 la ham sd chan => dpcm. Vi du 9: Xac dinh tam ddi xung cua dd thi mdi ham sd sau day: 3x-2 ,, 3x2 -5x + 5 a) y = b) y = x + 1 x- 2 Giai a) Dd thi cd TCD: x = -1, TCN: y = 3 nen cd giao didm I(-l; 3). Chuyen he 7^ fx = X-l true bang phep tinh tien vecto OI: { ly=Y+3 The vao ham sd: Y + 3 = 3(X-1) ~2 co Y = — Vi Y = F(X) = — la (X-l ) + l X v ; X ham sd le ndn dd thi ddi xung nhau qua gdc I(-l; 3). b) Tacdy= 3x' ~5x + 5 = 3x + 1 -nen cd TCD: x = 2, TCX: v = 3x + x r 2 t x- 2 1 va cd giao didm 1(2; 5). Chuyen he true bang phep tinh tidn theo vecto OI:jX=X+2 [y=Y+5 The vao ham so: Y + 5 = 3(X + 2) + 1 co Y = 3X - — la (X + 2)- 2 X ham sd le => dpcm. Vi du 10: Tim m de duong thang y = -x - 4 cat dd thi ham sd y = x +(m + 2)x—m ^. ^ (j-m j Xung nhau qua dudng thang y = x. x + 1 Giai Dieu kidn PT HDGD cd 2 nghiem phan biet khac -1: i^l2 ^ =_x-4co2x2 + (m + 7)x + 4-m = 0 (1). x + 1 -BDHSG DSGT12/1- 175 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
DK: |2(-l) 2 - ( m + 7) + 4-m* 0 [A = (m + 7) 2 -8(4-m)>0 m*- 1 m<-ll-V10 4 hay m>-llWl0 4 Goi xi, X2 la hoanh do hai giao diem, ta cd Xi, x2 la nghiem cua (1) theo dinh li Viet: x, + x. m + 7 Hai giao diem doi xung qua dudng thang y = x vuong gdc vdi dudng thang y = - x - 4 nen tung do cua hai giao diem lan luot la x2 , Xj. Do do X2 = -X|-4oxi+X 2 = -4o m + 7 = 8om = l (thoa man). V | du 11: Tim hai diem A, B thudc dd thi ham sd y = — nhau qua duong thang d: y = x + 3. Giai Xet dudng thang d' vudng gdc vdi d thi d': y = - x + b. •2x + 2 x - l ddi xung PT HDGD cd d' va (C): x -2 x + 2 x - l = - x + b, x * 1. co 2x2 - (b + 3)x + 2 4- b = 0. Didu kidn A = (b + 3) 2 - 8(2 + b) = b2 - 2b - 7 > 0. Hoanh do giao diem I ciia d va d': b - 3 x4-3 = - x + b =>x : = , X "f" X I la trung diem doan AB: Xi = — b - 3 b4-3 , . co = co b = 9 (chon). Vay A 4 14 6 + B X + X 4- 2 V i du 12: Tim hai diem E, F thudc dd thi ham sd y = : — ddi xiing x - l nhau qua didm 1(0; — ). Ta co y = x + 2 + Giai Goi E(xi; yi), F(x2; y2) theo dd bai: x 1 + x 2 = 0 x - l x1+x2=0 4 4 Xj 4-Xj 4-44 +- Xj + Xg = 0 X - l Xj " I Do do Xj =-x2, x2 =-9 nenE(-3;-2) vaF(3; 7). 176 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Vi du 13: Tim m de do thi ham sd y = x + 2m x + m ^ ^ ^.^ m x + 1 biet ddi xung nhau qua gdc O. Giai Hai diem thudc dd thi ddi xung nhau qua gdc O cd toa do (x; f(x)) va (-x; f(-x)). Xet phuong trinh f(-x) = -f(x), x * 0. x 2 -2m 2 x + m2 x 2 +2m 2 x + m2 , x * 0, x * ±1. - x + 1 x+ 1 co (2m 2 -l)x 2 = m2 , x * 0, x *+1 (1). Dieu kidn can tim la phuong trinh (1) cd 2 nghiem phan biet: 2m2 - 1 > 0, m2 * 0, m2 * 2m2 - 1 co m2 > -, m2 * 0, m2 * 1 co m < —]= hoac m > ~, m * ±1. 2 y/2 42 Vi du 14: Cho f(x) la ham da thuc bac 4. Chung minh dd thi cua f(x) cd true ddi xung x = a khi va chi khi: f'(a) = 0 vaf"(a) = 0 Giai Ta khai trien f(x) theo x - a: f(x) = a4(x - a) 4 + a3(x - a) 3 + a2(x - a) 2 + a^x - a) + a 0 f(i) (a ) trong do: aj = — — ndn dd thi cua da thuc f(x) cd true ddi xung x = a i ! khi va chi khi g(x) = f(x - a) la ham sd chan: ' f(3) ( a ) [ a 3 = 0 [ a 1 = 0 = 0 3! ~ " ff'(a) = 0 f(i)( a ) °|f"(a ) = 0 1! • Md rdng cho cac da thuc bac chan 2m ma dd thi cd true ddi xung x = a khi va chi khi f '(a) = f '"(a) = = fm ~] \a) = 0. Vi du 15: Tim quy tich cac tam ddi xung cua dd thi: a) y = x3 - 3mx2 + 4x - m2 + 1. 4x2 + 3(1 - 3m)x + 1 + 2m2 - 6m b) y = x - 2m Giai a) Tam doi xung la diem udn cua dd thi: y' = 3x2 - 6mx + 4, y" = 6x - 6m, y" = 0 CO x — m. Didm udn I co toa dd x = m, y = f(x). Khu m bang each the m = x vao y thi y = x3 - 3x3 + 4x - x2 + 1 = -2x3 - x2 + 4x + 1. Vay quy tich cac tam ddi xung la duong cong: y = -2x3 - x2 + 4x + 1 -BDHSG DSGT12/1- 177 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
b) Tam doi xung la giao diem 2 tiem can. y = 4x + 3 - m + — • x # 2 m x-2 m nen TCD: x = 2m, TCX: y = 4x + 3 - m, do do giao diem tiem can I cd toa do x = 2m, y = 4x + 3 - m. Khu m bang each the m = — vao y thi ti v = 4x + 3 - - = -x + 3. Vay quy tich cac tam doi xung la dudng 2 2 7 thang y = - x + 3. ti V i du 16: Tim quy tich cua diem: . , , , , , . x 2 -2m x + 3m - 5 a) Cue dai cua do thi y = x - l b) Cue tieu cua do thi y = x3 + 3mx2 + 3(m2 - l)x + m3 - 3m. Giai s ^ ^. v r., x 2 -2m x + 3m- 5 x 2 -2mx- m + 5 a) D = R \ {1}, y' = ; = = , A = m - 4. (x-l) 2 (x-l) 2 Dieu kien cd 2 cue tri la A > 0, 1 - 2 - m + 5 * 0 co m > 4, hoanh do cue tri la x = 1 ± Vm - 4 . Lap BBT thi diem cue dai A: x = 1 - Vm - 4 , y = f(x). Ta cd x = 1 - Vm - 4 => m = 4 + (1 - x) 2 the vao y thi duoc quy tich cac diem cue dai thuoc (P): y = -2x2 + 6x - 10, v i x = 1 - Vm - 4 < 1 nen gidi han la x < 1. b) D = R, y' = 3x2 + 6mx + 3(m2 - 1). Vi A' = 9 > 0, Vm nen do thi luon cd CD, CT cd hoanh do x = - m ±1. Lap BBT thi diem cue tieu la B: x = 1 - m, y = f( l - m) = -2. Vay quy tich ciia diem cue tieu la dudng thang d: y = -2. V i du 17: Vdi cac gia tri nao cua m dudng thang y = m - x cat do thi 2x2 - x + 1 tai hai diem phan biet A, B. Tim tap hop cac trung diem x - l M cua doan AB do. Giai 9y2 — y X 1 PTHDGD: = m - x <=> 3x2 - (m + 2)x + m + 1 = 0, x * 1. x - l V i x = 1 khong phai la nghiem nen dudng thang cat dudng con da cho tai hai diem phan biet khi va chi khi: A = (m + 2) 2 - 12(m + 1) > Oco m2 - 8m - 8 > 0 o m < 4 - 2 V6 hoac m > 4 + 2 V6 178 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
- A XA +X, m + 2 Hoanh do trung diem M ciia doan thang AB la: x M = 2 6 Vi diem M nam tren dudng thang y = m - x nen y M = m - x M . Khu m, ta cd m = 6x M - 2 nen y M = 6x M - 2 - x M = 5x M - 2. Vay tap hop cac diem M la nam tren dudng thang y = 5x - 2. Gi di han: m < 4 - 2 V6 ^ 6x - 2 < 4 - 2 76 co x < 1 - ^ va m > 4 + 2 76 3 =>6x-2> 4 + 2V6ox> l 3 Yj_du_18: Chung minh cac dd thi y = x3 + mx2 - m - 1 ludn di qua 2 didm cd dinh A, B. Tim quy tich cac giao diem cua 2 tidp tuyen tai A va B. Giai Goi M(x; y) la diem cd dinh ciia cac dd thi: y 0 = xf+mxf - m - l , Vmoy 0 = m(xf - 1) + xf - 1, Vm. x o=- T y 0 = - 2 x 0 = l , y o = 0 Vay cac do thi qua 2 didm co dinh A(l ; 0) va B(-l ; -2). Ta cd y' = 3x2 + 2mx nen 2 tidp tuyen tai A va B la: y = (2m + 3)(x - 1), y = (-2m + 3)(x + 1) - 2. Khu m thi quy tich cac 3x2 —Y — 9 giao diem la dudng cong y = — - x V l d u 19 : Tim quy tich cua didm thudc true tung ma tir do ve it nhat mdt tiep x2 — x + 1 tuyen vdi dd thi y = x - l Giai x(x — 2) Ta cd D = R \ {1} , y' = _ ndn phuong trinh tidp tuyen tai diem M \x — 1) co hoanh dd x 0 * 1. y = ^- 2 ) (x-x o ) + ii ^ Chox = 0thi y 2x - X (*.-!) ' x„ - 1 Xet f(x) = ^— L x * i thi f'(x) = -^ x - (x-1) 2 (x-l) 3 Chof'(l) = 0co x = 0. -BDHSG DSGT12/1- 179 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Bang bien thien X -oo 0 +00 y' - o + - y +00 0 Do do y > -1 , nen quy tich cua diem thuoc true tung can tim la B(0; y) vdi y > 1. V i du 20: Tim quy tich cac diem ma tii do ve dugc 2 tiep tuyen den (C): y = x - 1 + - - ma 2 tiep tuyen nay vuong goc vdi nhau. x - l Giai Goi M(a; b), phuong trinh ducmg thang d qua M cd he sd gdc k: y = k(x - a) + b. Dieu kien d tidp xuc (C) la he sau cd cd nghidm x * 1 1 -- Do do x — 1 H — - — = k(x - a) + b x - l 1 <=> < X - 1 + x-1- 1 x - l 1 = k(x - a) + b k(x-l) i ) k(l-a ) + b^> l--(k(l - a )^T)) 2 = k. x - l 4 Ta cd phuong trinh bac 2 theo he sd gdc k: g(k) = (a - l) 2 k 2 + 2((1 - a)b + 2)k + b2 - 4 = 0, k * 1. Ydu cau bai toan: a * 1, k]k2 = -1 , g(l) * 0. o a * 1, b2 - 4 = -(a - l) 2 , (a - l) 2 + 2((1 - a)b + 2) + b2 - 4 * 0. co (a - I) 2 + b2 = 4, a * 1, a * b + 1. Vay quy tich cac didm can tim la duong trdn (x - l) 2 + y2 = 4 bd di 4 diem A(l ; 2), B(l ; -2), C(l + -J2 ; 72) va D(l-V 2 ;-&). V i du 21: Tim quy tich cac diem ma tir dd ve dugc 2 tiep tuyen ddn (P): y = ax2 + bx + c, a * 0 va 2 tiep tuyen nay vudng gdc nhau. Giai Goi M(x0 ; y0 ), phuong trinh duong thang d qua M cd he sd gdc k la: y = k(x - x0 ) + yG. Jax2 +bx + c = k(x-x o ) + y0 Dieu kidn d tiep xuc (P) la he sau cd nghiem: 2ax + b = k 180 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
<=> < X = k - b 2a k - b 2a + b k - b + c k - b - x„ + y 2a ) \ 2a Ta dugc phucmg trinh bac 2 theo he sd gdc k. k 2 - 2(b + 2ax0 )k + b2 - 4ac + 4ay0 = 0 (1). Dieu kidn ve dugc 2 tiep tuyen vudng gdc la A > 0, P = - 1 „ , , , „ b2 -4ac +1 o r = -l o -b + 4ac - 4ay0 = - 1 <=> y0 4a Vay quy tich can tim la dudng thang cd PT: y = — - 4ac + 1 4a Ket qua: Dieu kidn ve dung 1 tiep tuyen la phuong trinh (1) cd cd nghiem kep: A = 0 nen quy tich chinh la parabol: y = ax2 + bx + c. C. BA I LUYEN TAP Bai 1: Tim m de dudng thang: y = mx + x - 1 cat (H): y = thudc mdt nhanh dd thi. DS: m<0 , m *- 3 Bai 2: Tim m de dudng thang y = mx + 1 cat do thi y x + 2 2x + l tai 2 diem x + 1 x - l tai hai diem thudc 2 nhanh. DS: m> 0 Bai 3: Cho ham sd y = x 3 - 3x2 + 4. Chung minh moi dudng thang qua 1(1; 2), he sd gdc k > -3 deu cat dd thi tai 3 diem phan biet I , A, B ddng thdi I trung didm AB. Bai 4: Tim m de dd thi: y = mx3 - x2 - 2x + 8m cat true hoanh tai 3 diem phan biet. n c 1 1 DS: — < m < — 6 r 2 Bai 5: Cho ham sd y = x3 - 3m + 2. Goi d la dudng thang qua A(3; 20) cd he sd gdc m. Tim m de d cat dd thi tai 3 diem phan bidt. 15 m > DS: \ 4 m * 24 Bai 6: Tim m dd dd thi: y = — x3 - mx2 + (m2 - l)x - — m3 cat true hoanh tai 3 3 3 diem phan bidt. -BDHSG DSGT12/1- 181 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
9 9 DS: - - <m < - 3 3 Bai 7: Tim m de do thi: y = x4 - (3m + 4)x2 + m2 cat true hoanh tai 4 diem phan biet cd hoanh do len cap cong. 12 DS: m = hoac m = 12 19 Bai 8: Tim m dd db thi: y = x4 - 2mx2 + m3 - m2 tiep xuc vdi true hoanh tai 2 diem phan biet. DS: m = 2 Bai 9: Chung minh 2 do thi y = x3 - x va y = x2 - 1 tiep xuc nhau. —x2 + 2x — 3 Bai 10: Chung minh 2 dd thi y = x3 - 3x - 1 va y = tiep xiic x - l nhau. Lap phuong trinh tiep tuyen chung. DS: y = x- 5 Bai 11: Viet phuong trinh tiep tuyen cua dd thi ham sd: y = 4x3 - 6x2 + 1, biet tiep tuyen di qua M(-l ; -9) DS: y = 24x+15 vay = —x- — 4 4 x 2 + x - 1 Bai 12: Cho ham sd: y = Lap phuong trinh tiep tuyen ciia dd thi, x + 2 bidt tiep tuyen vudng gdc vdi tiem can xien. DS: y = -x ± 2^2 - 5. Bai 13: Cho ham so: y = — x3 - 2x2 + 3x. Lap phuong trinh tiep tuyen cd he sd gdc be nhat. DS: y = - x + 3 1 3 m 2 1 Bai 14: Cho ham sd: y = — x x +— Goi M la diem thudc do thi cd hoanh • ^ 3 2 3 ' ' do x = -1 . Tim m de tiep tuyen tai M song song dudng thang: 5x - y = 0. DS: m = 4 Bai 15: Chung minh qua M(0; 1) cd 3 tiep tuyen ciia ddthi(C):y=-x4 -2x2 +l Bai 16: Chiing minh cac do thi cua ham sd: y = x4 - (m + l)x2 + m ludn di qua 2 diem cd dinh vdi moi m. DS: (-l;0)va(l;0 ) ' +i - m X ~~ 1 _L1 Bai 17: Tim diem co dinh cua cac do thi: y , m * +1 x - m DS: (l;-l)va(-l ; 1) 182 -BDHSG DSGTUIlDownload Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Bai 18: Chung minh vai moi m thi duong thang y = mx + m - 1 ludn di qua x + 2 1 diem cd dinh thudc (H): y = 2x +1 DS: A(l ; -1) thudc (H) Bai 19: Chung minh 2 dd thi y = 2mx - 4m + 3 va y = x3 - 3mx2 + 3(2m - l)x + 1 cd mdt diem cd dinh chung vdi moi m. DS: A(2;3). Bai 20: Chung minh cac dd thi: y = x3 + (m - l) x 2 - 2(m + l)x + m - 2 tiep xiic nhau tai 1 diem cd dinh. DS: tai diem Mo(l; 4) va tiep tuyen chung y = - x + 5. Bai 21: Chung minh cac dd thi: y = 2x + mx + 2 —™ m ^ 2 lu5n tiep xuc x + m - 1 vdi 1 dudng thang cd dinh tai 1 diem cd dinh. DS: dudng thang y = x - 1 cd dinh tai diem cd dinh (-1 ; -2). Bai 22: Tim cac diem thudc mat phang ma cac dd thi: y = x3 - mx2 + x + m khdng di qua vdi moi m. DS: 2 dudng thang x = ±1 bd di 2 diem (1; 2) va (-1 ; -2). Bai 23: Chung minh tren (P): y = x2 cd 2 diem ma khdng thudc dd thi y = 2x3 - 3(m + 3)x2 + 18mx - 8 vdi moi m. DS: 0(0; 0), B(6; 36) 2x2 - x + 1 Bai 24: Cho ham sd y = Tim m de dudng thang d: y = m - x cat x - l dd thi tai 2 diem A va B. Tim tap hop trung diem I cua AB khi m thay ddi. DS: y = 5x - 2 vdi x < 1 - --^ hay x > 1 + , 3 3 Bai 25: Tim tap hop cac diem udn cua dd thi: y = 2x3 - 3(m - 2)x2 - (m - l)x + m khi m thay ddi. DS: y = -4x 3 -2x 2 + x + 2 Bai 26: Tim tap hop cac tam ddi xung cua dd thi: y = m x + — khi m thay x + m - 2 ddi, m * -1 ; 3 DS: y = - x + 2 tru 2 didm (3;-1) va(-l ; 3). Bai 27: Tim tap hop cac diem cue dai cua dd thi: y = — 3x + m khi m x - 2 thay ddi. DS: y = 4x - 3 vdi x < 2 ^ 1. 1 ax 2 +(2 x + l)x + a + 3 , A Bai 28: Cho ham so: y = , a * -1 , a * 0 x + 2 Tim a dd 2 tiem can hop nhau gdc 30° DS: a= ±S -BDHSG DSGT12/1- 183 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
x2 — 3x Bai 29: Cho do thi (C): y = - — — cd tam ddi xung I . Tim m de dudng x - l thang d: y = m - x cat dd thi tai 2 diem A, B ma IA vudng gdc IB. DS: m = ±2 „,.,« r. i. x2 +2(m + l)x + m2 +4m „, , , i . Bai 30: Cho ham so y = Tim m de ham so co cue x + 2 dai va cue tieu, ddng thdi cac diem cue tri cung vdi gdc toa dd tao thanh tam giac vudng tai O. DS: m = -4±2V6 Bai 31: Tim m dd dudng thang d: y = 1 - x cat do thi y = x3 + mx2 + 1 tai 3 diem phan bidt A(0; 1), B va C ma tiep tuyen tai B va C vudng gdc nhau. DS: m = - — 9 Bai 32: Cho ham sd y = mx + — Tim m dd ham sd cd cue tri va khoang each tir diem cue tieu den tiem can xien bang V2 DS: m= 1 2x ' Bai 33: Cho ham sd y = Tim toa do M thudc dd thi, biet tiep tuyen tai x + 1 M cat 2 true toa do tai A, B va tam giac OAB cd dien tich bang — 4 DS: M(l;-l)v M(-^;-2) Bai 34: Cho ham sd y = -x3 + 3x2 + 3(m2 - l)x - 3m2 - 1. Tim m de ham so cd cue dai, cue tieu va cac diem cue tri each deu gdc O. DS: m = ±- 2 Bai 35: Cho ham sd y = x +(m + l)x + m + l minh vdi moi m, dd x + 1 thi ludn cd cue dai, cue tieu va khoang each giua 2 diem do bang V20 DS: MN= V20 x 2 + l Bai 36: Cho ham sd y = Tiep tuyen tai diem M thudc dd thi cat tiem x can dung tai A, tiem can xien tai B. Chung minh M la trung diem AB va S(OAB) khdng phu thudc vi tri M. 2x +1 Bai 37: Tim diem M thudc dd thi y = cd tdng khoang each ddn: x + 1 a) 2 ti?m can be nhat. b) 2 true be nhat. 184 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
DS: a) M(0; 1), M(-2; 3) Bai 38: Chung minh tich khoang each tir M tuy y thuoc do thi den 2 tiem can la mot so khdng ddi: x 7x - 2 , N x 2 +3x- l a) y= b) y = 5x + 4 ' J x- 2 DS: b)T = A ; khdng ddi V2 Bai 39: Tim tap hop cac diem trong mat phang ma tir dd ve duoc it nhat mdt tiep tuyen den do thi: 3x4-5 x 2 + 3x4 - 4 a) y = b) y = 2x- l x - 2 Bai 40: Tim tap hop cac diem trong mat phang ma tir do ve duoc it nhat mdt tiep tuyen den dd thi: a) y = 3x2 + 6x - 32 b) y = x4 - 5x2 + 1 Bai 41: Tim tap hop cac didm trong mat phang ma tir do ve dung 2 tiep tuyen den dd thi: x - 6 , , -2x 2 -x4 - 9 a) y = b) y = — — x + 3 3x - 2 Bai 42: Tim tap hop cac diem trong mat phang ma tir do ve dung 2 tiep tuyen den dd thi: a) y = 4x2 - 8x - 1 b) y = 3x4 + 7x 2 - 12 Bai 43: Tim tap hop cac diem trong mat phang ma tir dd ve dung 2 tiep tuyen ddn do thi va 2 tiep tuyen nay vudng gdc nhau: 3-4 x l x 2x 2 +4x-1 3 a) y = b) y 4x+3 x+ 3 Bai 44: Tim tap hop cac diem trong mat phang ma tir do ve dung 2 tidp tuyen den dd thi va 2 tiep tuyen nay vudng gdc nhau: a) y = 2x2 - 13x + 5 b) y = 4x4 - 2x2 + 7 x2 + 6x + 1 Bai 45: Tim 2 didm tren 2 nhanh dd thi y = cd khoang each be nhat. x - 4 x - 2 Bai 46: Tim 2 diem tren 2 nhanh dd thi y = cd khoang each be nhat. x - l DS: A(0; 2), B(2; 0). -BDHSG DSGT12/1- 185 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
CHUON G II : HA M S O LU Y THUA , HA M S O M U V A HA M S O LOGARI T § 1 . QU Y TA C BIE N DO I V A CA C HA M S O A. KIEN THUC CO BAN • Luy thua va can thuc: - Luy thua vai so mu nguyen duong: a" = a.a...a, n thua sd a (vdi moi a vaneN ) - Luy thua vdi sd mu 0 va nguyen am: a 0 = 1 va a _ n = (vdi a # 0 va 9. n e N*) - Luy thua vdi sd mu hiiu ti; ar = a" = \jam a (vdi a > 0 va r = —, n e Z, n n e N¥ ) - Luy thua vdi sd mu thuc: aa = lima 1 " (vdi a > 0, a e R, r n e Q va limr n = a). - Can bac n: Khi n le, b = yfa <=> b" = a ( vdi moi a) fb > 0 Kh (vdi a > 0). i n chan, b = va <=> b n =a - Bien ddi luy thua: Vdi cac sd a > 0, b > 0, a va P tuy y, ta cd: a<\ap = aa+p ; aa : a p = aa ~p (a a ) p = a a p (a.b) a = a" b a ; (a: b) a = aa : b a - Quan he so sanh: Ndu a > 1 thi: aa > a p » a > p Neu 0 < a < 1 thi: a°>a p oa< p Neu 0 < a < b thi: a°<b a oa>0,a a > b o l oa<0 . - Bien ddi can bac cao: Vdi hai sd khdng am a, b, hai sd nguyen duong m, n va hai sd nguyen p, q tuy y, ta cd: %/ab = Va.\/b ,J=^(b>0 ) ,vaP=(va f (a>0);# T = mv/a" b jyb Neu ^ = 2L thi Va p = ^a q (a > 0). Dac biet Va = m ^ a m n m • Logarit: - Logarit ca sd a: a = logab o a" = b (0 < a * 1 va b > 0) - Logarit ca sd 10: logiob = Igb hay logb 186 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
- Logarit co so e: logeb = lnb (e « 2,7183) - Tinh chat: loga l = 0 va loga a b = b vbi a > 0, a * 1. a i°g.b =bvaia>0,b>0,a*l . - Bien doi logarit trong dieu kien xac dinh: loga (b.c) = logab + logaC l0g a - = l0gab - logaC , log: C "logaC logaba = cxlogab (vdi moi a), loga Vb =-log a b (n e N") n - Ddi eo sd trong dieu kien xac dinh: log a x logbX = logba = logab 1 hay logab . logb x = logax hay logab.logba = 1; tog . a logab log a b - Quan he so sanh vdi a > 0, a * 1, b > 0, c > 0. Neu a > 1 thi: logab > loga c <=> b < c. Neu 0 < a < 1 thi: logab > loga c o b < c. Neu a > 1 thi: logab >0ob>l . Neu 0 < a < 1 thi: logab >0ob<l . logab = logaC <=> b = C. • Cdng thuc lai kep, tang trudng mu: Gui tien vao ngan hang theo the thuc lai kep theo dinh ki: neu den ki han ngudi gui khdng rut lai thi tien lai duoc tinh vao vdn cua ki ke tidp. Neu mdt ngudi gui sd tien A vdi lai suat r mdi ki thi sau N ki sd tien nguoi ay thu duoc ca vdn lan lai la: C = A( l + r) N Gia sir ta chia moi ki thanh m dot de tinh lai va giu nguyen lai suat mdi ki la r thi lai suat moi dot la — va sd tien thu duoc sau N ki (hay sau Nm m .Nm dot) la A 1 + - m The thuc tinh lai khi m -> +QO goi la the thiic lai kep lien tuc. Nhu vay vdi so vdn ban dau la A, theo the thuc lai kep lien tuc, lai suit moi ki la r thi sau N ki sd tien thu duoc ca vdn lan lai se la: S = Ae N r , duoc goi la cdng thuc lai kep lien tuc hay tang trudng mu • Ham so luy thira y = x": Lien tuc tren tap xac dinh cua nd Dao ham (xa )' = ax™-1 , (ua )' au^u': (rfc\=—L= (x>0), (vu) = , vdi u = u(x) > 0. 1 ' nVx 1 ^ _ _ 1 ' nv ^ Ham sd y = x a ddng bien tren (0; +oo) khi a > 0; nghich bidn tren (0; +<x>) khi a < 0. -BDHSG DSGT12/1- 187 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
• Ham so mu: Lien tuc tren tap xac dinh R, nhan moi gia trj thuoc (0; +00). [+cokhia > l r (0 khia> l hm a" = < ; hm a = < ^-> +* [0 khiO<a< l *-»-« [+ookhi0 <a< l Dao ham: (a x )' = n = ax lna; (e x )' = ex ; (a u )'= aVlna; (e u )'= e V vdi u = u(x). f Ddng bidn trdn R neu a > 1, nghich bien tren R neu 0 < a < 1. Dd thi ludn cat true tung tai diem (0; 1), nam a phia tren true hoanh va nhan true hoanh lam tiem can ngang. • Ham sd logarit y = logax: Lien tuc tren tap xac dinh (0; +00), nhan moi gia tri thudc R. , [+00 khi a >1 ,. , [-00 khi a >1 hmlogax = ^ ; limlog x = < X-VKO [-ookhi0<a<l x-++o+ [+ookhi0<a<l 1 1 • 1 Dao ham (log x) = ; (In x)' = — ; (in Ixl) = — x i n a x v 1 u x (logau)'= ;(lnu)'= — ;(ln|u|)'=— vdiu = u(x). uln a u u Ham sd y = logax ddng bien tren (0; +00) neu a < 1, nghich bien tren (0;+oo)ndu0<a< 1. Dd thi ludn cat true hoanh tai diem (1; 0), nam d ben phai true tung va nhan true tung lam tiem can dung. • Dang vd dinh: hm 1 + =e; hm-—- = 1 ; lim^n ^ + X^ =1 x-+*c{ xj X *-»0 X x->0 B. PHAN DANG TOA N DANG 1: BIEN DOI LUY THUA - MO - LOGARIT - M d rdng luy thua: Va, n e N*; a" = a.a...a ( n thira sd) a*0 , n e N*:a°= l,a~" = — a > 0, n e N*, m e N*: a "~ = \[a™ , a~» = — a n - Sir dung cac bien ddi ve luy thira, can bac n. - Vdi a > 0, a * 1: a M = N co M = logaN r - Sir dung cac bien ddi logarit, ddi co sd. Chuy : • 0° vd nghia. x . Vx * x 3 (do didu kien xac djnh khac nhau). 188 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
• Va i s6 x > 1, trong each viet trong he thap phan thi so chu so dung trudc dau phay la n+l, trong do n la phan nguyen ciia x: vi 10"<x < 10 n+1 »n<logx< n + 1 nen n = [logx]. . Cong thiic lai kep C = A(l + r) N Vi du 1: Thuc hien phep tinh i A = 81^ 7 5 + f 1 1 -a r i T 1125 J U J B = 0,001 3 -(-2)" 2 .643 - 8 3 +(90 ) 2 C = 273 + •25u D = (-0,5) -625° 1) 2 + 19(-3)~ (3) = (3)" _3 ( 4 \ 4 3 Y Giai _3 J V 1 'IV 3 l 1 „ 80 — + 5- 8 = 3 = 27 27 27 B = (10" V -2" 2 .(2 6 ) 3 -(2 3 ) 3 +1 10- 2 -27* +1 = 7 1 111 16 16 C= (33 ) 3 +(2" 4 )" 4 -(5 2 ) 2 = 32 + 23 - 5 = 12 D=((-2)-f-(5^ - V 2 2 2 19 + • -27 2 4 -5. | - 2 l i = n_JL_ H = io . 27 27 27 Vi du 2; Tinh gia tri cac bieu thiic sau: E= (0,5^)^; G = 3 1+2 ^ : 9^2 ; 0 , 5 ^ ) 7 i =0,5^=0 > 5 4 =(i[= ^ H = (4 2 7 3 -4S - l ).2~2S Giai 4 16 p _ 22-3%/5 g\/5 _ 22-3V5 23 V5 _ 22-3%/5+3%/5 _ 22 = 4 • G = 3 1+2*2 9*2 =3 1 + 23/2 . n2?/2 = 3 1+242-2*2 n3l = 3 H = (4 2J3 _ 4^3-!) 2 - 2 ^ = 24 ^ 2~2 ^ 3 -22 v ' 3 " 1 2~2v/3 -2 2 ^ —— -BDHSG DSGT12/1- 189 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
V i du 3: Viet cac bieu thuc sau dudi dang luy thua cua mot so vdi sd mu huu ti: I = vx ^ W (x > 0); j = 5fe 3 p ( a >0\b>0 ) V a V b -mil - ( ( i M f i\i ? = VaVaVaTa" : a 1 6 (a > 0) Giai x V X V I .12 J = K = L 1^5 ( a )3 b 1 2^5 a 1 3 b v 1 N 1 iir r 2 1 3 6 1 1 1.^3 ( 2 > a i is b I. 2^2 f 1 1 1 1 \ 11 15 11 V i du 4: Don gian bieu thuc trong didu kien xac dinh: , , Va - Vb Va + Vab . T a - b a + b M=^ = == = ^ : N : /a + v / b Va"-Vb Va" + Vb i i P = M 7 a - 1 Va + Va 7 . ~ a 3 - a 3 a 3 - a ; /a + 1 .a4 +1 ; Q Giai (Va" + Vb)(Va"-Vb) Va"(Va~ + Vab) V^-V b Va" + Vb N = Va T + Vab + Vb T -(Va T -Va ^ + Vb T ) = 2Vab = Va + Vb-V a = Vb (Va + l)(Va" -1 ) Va(Va" + 1) Va(Va+l) (Va + 1) / a + 1 = v a - 1 + 1 = V a ajq-a 2 ) a ~ ^(l-a 2 ) n n , - Q - -17 in = (1 + a) - (1 - a) = 2a. a / 3 (l-a ) a (a + 1) V i du 5: Rut gon cac bieu thuc: a) R Vax3 - 4 / - 3 a x 1 + vax + ; la - Vx /ax ! ^ x + x ^ S > ° ' X > ° ' a 190 -BDHSG DSGTU/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Giai a) Ta co - 4 / a x 7 - Va^x + 1+yax = -Vax:(7a - y/x) + 1 + Vax a-V x vax Va - V x 1 + Vax -Vax + 1 + Vax 1 /ax -Vax + /ax i + /ax = 1 + /ax Do clo R = Vax 1 + . * = Va(vx + Va) b) Datu a +Va 2 - b a-Va 2 - b (u > v > 0) thi u 2 + v 2 = a; u2 v2 = — nen b = 4u2 v2 nen 4 / TTvf l~2 2 o a +Va 2 - b a-Va 2 - b Va + Vb=V u + v + 2uv = u + v = , + « Tuong tu: %/a-Vb = Vu2 +v 2 -2uv =u - v = J a + a - - J a Vi du 6: Chung minh a) V4 + 2V3 - V4-2V3 =2 b) V9 + 780 + 79- 780 = 3 Giai a) Vi V4 + 2V3-V4-2V3 >0 nen >/4 + 2V3 - V4-273 =2 o (V4 + 2V3-V4-2V3)2 =4 o 4 + 2V3 + 4 - 2V3 - 2Vl6 -12=4 (dung). Cach khac: Taed 4 ±2 73 = (73 )2 ±2 73 + 1 =(S ± l)2 b) Dat x = 79 + V80 + 2/9 - V80 Taco: x3 = 9+ V80 + 9-^+33 i9+V80.3 /9-x/&0(3 /9 + ^ = 18 + 3781-80.x = 18 + 3x . Do do co phuong trinh: x3 - 3x - 18 = 0 <=> (x - 3)(x2 + 3x + 6)ox = 3o dpcm. Cach khac: 3±S ) 72 ±3275 r —7,— = ~ = 9 ± 475 = 9 ± •BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
nen ^9 + VSO + ^9 - Vso = 3 + ^ + 3 ~ ^ = 3 2 2 Chii y: Co the diing S = 3, P = 1 de tim nghiem cua X 2 - 3X + 1 = 0. V i du 7: Khdng dung may, tinh gia tri dung: a) Vl5 + 6V6 + Vl5-6V 6 b) 77 + 5V2 - ^7 - 5 ^ 2 Giai a) Tacc(3V2 ±2V3) 2 = 18 +12 +12^ 6 =30 +12 ^ 6 nen ylT^Tf + V^yj = 3^ JL2 ^ + 3^ ~12 ^ 3 = 6 V2 V2 Cach khac: Dat yJ15 + &j6 + Jl5~6S = x, x > 0. Ta cd x2 = 30 + 2 ^225-216 = 36 nen chon x = 6. b) Taed: 7 + 5V2 = 1 + 3V2 + 6 + 2V2 = (1 + V2) 3 Tuong tu 7 - 5 s/2 = (1 - V2 ) 3 Do dd ^7 + 572 - ^7-5^2 = 1 + V2-(1-V2 ) = 2 ^ Cach khac: Dat x = v/ 7 + 5v/ 2 - -5^2 Ta cd: x3 = 7+5 V2 - (7-5 72 )-3(^7-5v/2 - 3 /7 + oV2 ).(^(7+5>/2)(7-5v/2)) = 10^2 + 3(^7 + 572 - ^7 - 5V2) = 10V2 + 3x. Ta cd phuong trinh: x 3 -3x-10v / 2 =0co (x-2v^)(x 2 + 2V2 x + 5) = 0cox = 2v/ 2 V i du 8: a) Cho x > 0, y > 0, hay bieu thi x qua y biet rang: 2 _1 _3 y = x 3 , y = 2x 4 , y = x 2 -1 . 1 1 b) True can 0 mau V 2+ 3/3 5-^13 Giai 48 2 3 _i_ , \-4 a) y = x 3 =ox=y 2 ;y=2x 4 =ox= P _3 _3 _2 y = x 2 - 1 => x 2 = y + l=> x = (y + l) 3 1 1/3-42 ( 3 y3-V2)(3^ 3+ 2 3 / 9 + 4) b) 72 + ^3 1/9-2 1 V i 5 - V13 + V48 =5-7(2^3 + 1)2 =4-2^ 3 =(V3-2) 2 192 -BDHSG DSGT12/ Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
nen ^-l) 2 _ (73 + 1)^4-273 2 Vi du 9: a) Cho x > 0, y > 0, z > 0. 2 2 2 Chiing minh: x 3 + y 3 = z 3 co (x2 + y2 - z 2 ) 3 + 27x2 y2 z 2 = 0 a*-a ' b) Cho sh(x) ch(x) = 8 * a — ; th(x) = -— vdi a > 0, 2 ' x ' 2 a * 1. Chiing minh ch2 (x) - sh2 (x) = 1, th(2x) a + a 2th(x) l + th2 (x) Giai 2 2 2 a) Taed: x^3 _1_ T7 3 rj 3 3 +y 3 =z 3 <=> x 3 + y 3 = z^ 3 2 2 4 <=> x2 + 3x 4 y 3 +3x 3 y 3 + y2 = z 2 2 2 f 2 2\ X3 + y 3 V J o (x2 + y2 - z 2 ) 3 = -27x2 y2 z 2 C-> (x2 + y2 - z 2 ) 3 + 27x2 y2 z 2 = 0 => dpcm. co x2 + y 2 - z 2 = -3x3 y= 2 2 2 = r3x 3 y 3 z J b) ch2 (x) - sh2 (x) = a2x+a-2x+2_£ f a x +a- ^ 2 V-a" M I 2 J V 2 J 2x „-2x 4 4 \2 Taed: 1 + th2 (x) = 1 + 2th(x) a - a a + a" 2(a 2 x +a- 2 x ) a 2 x +a- 2 x + 2 nen = 2 a x -a" x a 2 x +a" 2 x + 2 a x +a- x 2(a 2 x +a' 2 x ) -2x 1 + th2 (x) _ 2(a x -a- x )(a x -a- x ) 2 _a 2 x - a - 2(a x +a- x )(a 2 x -a- 2 x ) a 2 x + a"2x Vi du 10: Cho sd tu nhien n le, chung minh: 1 1 1 1 a) Neu - + - + uv 1 1 a b c a + b + c a thi — + — + n bn c n a" + b n + c n b) Ndu ax" = by" = cz n , - + - + - = 1 thi: x y z -BDJ ifax*-1 + by"" 1 + cz""1 = 4a~ + 4b + 4c~ {SG DSGT12/1- 193 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Giai a) Tir gia thiet suy ra - + - = — a b a+b+c c => (a + b).(a + b + c)c = abc - ab(a + b + c) => (a + b)(b + c)(c + a),= 0 => co 2 so doi nhau ma ta cd n le => dpcm. , jaxn byn cz n | ~ ' b) VT = j> — —+^ — + = Wax V x y z V 1 1 1 _+. _ + _ x y z = Tax" = xTa" = yTb = zf c VT 1 1 D — + — + - x y z 7a + %/b + \/c => dpcm. V i dull : Tinh: a) log, 125 ; log^ ; lo g l — ; log, 36 5 ' 2 , 6 4 6 b) 3 log3ls ; 3 51oga 2 ;(-) a) log! 125 = log, 5 5 1 ^ v 3 3 Nlog2 5 f Y°go,52 32~J Giai "3; log05^ = log05 0,5 = 1 b) 31 4 4 531 8 = 18 ; 5 log, — = log, - =3 ; log, 36 = log, = -2 5 log3 2 _ = 3" = 2° =32 _/2~3 ) log 25 _2( ~3)log 25 = 2log 2 =5" 3 = 125 / y° g °' 5 2 v 3 2 j Vid u 12: Tinh: I 2 = 2 =32 a) log8 12 - log8 15 + log8 20 log5 36-log 5 1 2 b) -log7 36 - log7 14 - 31og7 </21 2 C) d) 36loge 5 +10 log5 9 •,l-log2 _ olog2 3 a) log8 12 - log8 15 + log8 20 = log8 — .20 = log8 42 = log2 3 2 b) Tog/ ^log7 36 - log7 14 - 31og 7 721 = log 7 [-^ - Giai (12 — 20 U s J ( 6 log7 14.21 c) log 5 36-log 5 1 2 _ log5 3 _ 1 log5 9 21og5 3 2 194 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
d) 36 ,og » 5 + l0 1 -loB2 -81og 2 3 = 6 loge52 +10 log ' ° 5 - 2 log23 3 = 52 + 5 - 33 = 3. Vi du 13: Tinh gon a) loga b) log5 log5 , n dau can o n i c)log72- 2 1og + .lo gVl0 8 d) log--log0,37 5 + 2log^0,5625 256 8 Giai: a) V t - V = a 3 5 4 = a 6 0 => log, \/a 2 3/7 5/„4 a'.Va.va 173 60 b) n dau can nen $Jl% =&6 > => log5 log5 = - n c) log 72-2 log— + logVi08 = log(23 .32 ) - log-^+logV ^ 256 2 = log 9I6 3> 23 .32 —.2.32 = 10g A 5 ^ 220 3 2 V ) = 201og2 — log3 2 d) log--log0,375+ 2log70^5625 = log2"3 - log(0,53 .3) + 21ogv'o,54 .3'2 8 3 = log2~3 - log2- 3 - log3 + 21og2"2 + 21og3 = log2"4 + log3 = log 16 Vi du 14: Rut gon cac bieu thuc: A = log36 log89 . log62 B = log32. log43 loge5 log76 log87 C = logab2 + loga2 b4 D = a^1 - b^1 Giai 112 A = log36.1og62.1og89 = log32 -log2 9 = -log 3 9 = - O O O B = log32.1og43 log54.1og65 log76 log87 log2 log3 log4 log5 log6 log7 _ log2 1 1 - . . . . . = iogR 2 = - log, 2 = - log3 log4 log5 log6 log7 log8 log8 8 3 2 3 C = logab2 + loga2 b4 = logab2 + ilogab4 = logab2 + logab2 = 21ogab2 2 Dat x = Jloga b => logab = x2 => b = ax ' Mat khac logba = —=> ^J\ogh a = —. Do du: D x x Vi du 15: Tinh logax, biet logab = 3, logac = -2 vdi: ax - a x = 0 a)x = a 3 b2 v^ -BDHSG DSGT12/1- b)x a 4 3 /b 19: Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Giai 0 logax = log a ( a 3 b 2 ^ ) = 3 + 21oga b + ilog a c = 3 + 2.3+ ^(-2 ) = 8 l 7b b) logax = loga V i du 16: Tim x biet: a) logx - = -1 = 4 + -log a b - 31oga c = 4 + - .3 - 3(-2) = 11. 3 3 b)logx 75 =-4 . c) log5 x = 21og5a - 31og5b 2 1 d) logj x = -log ! a + -log ! b 2 d 2 b 2 Giai a) log x i = - 1 o x"1 = - = 7"1 ci> x = 7. 7 7 b) logx S = -4 o x"4 = 75 o x = (75 I 1 ' 4 =5~8 a 2 a 2 c) log5 x = log5 a 2 - log5 b3 = log5 — ^> x = — 2 1 f 2 l\ d) logj x = log! a 3 + logi b5 = log1 a 3 .b£ 2 2- 2 2 V, V i du 17: a) Tinh log2 sl5 theo a = logis3. b) Tinh log41250 theo b = log2 5. c) Tinh log ^ 50 theo log315 = a, log3 10 = b. d) Tinh ln6,25 theo c = ln2, d = ln5. Giai a) log2515 = 2 1 x = a 3 .b5 log1 5 25 21og155 2aog1 5 15-log1 5 3) 2(1-a) b) log41250 = -log2(54 .2) = 21og25 + - = 2b + - c) log%/73 50 = log , 50 = 21og350 = 21og310 + 21og35 = 21og310 + 21og3 l^ = 21og310 + 2(log315 - 1) 3 = 2b + 2(a - 1) = 2a + 2b - 2. d) ln6,25 = ln(58 .0,5^ = 21n5 + 21n0,5 = 21n5 - 21n2 = 2d - 2c V i du 18: a) Cho log6 15 = x, logi 2 18 = y, tinh log2 5 24 theo x, y. b) Cho a = log2 3, b = log35, c = log7 2, tinh logi4o63 theo a, b, c. Giai 196 -BDHSG DSGT1Z/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
log, 3.5 log, 3 +log, 5 log, 2.3' l + 21og a) Ta co x = — ^ = — ^ ^— va y = 1 , = — log2 2.3 l + log2 3 log2 22 .3 2 + log2 c , - 2y- l . ,. x + l-2 y + xy Suy ra log2 3 = — ; log2 5 - 2 - y " 2- y .. Do do log2 5 24 = _ log2 2J .3 5- y log2 52 2(x + l-2 y + xy) b) log14063 = log140 (32 .7) = 2 log140 3 + log140 7 1 2 1 + • log3 140 log7 140 log3 (22 .5.7) log7 (22 .5.7) 2 1 = 1- 2 log3 2 + log3 5 + log3 7 2 log7 2 + log7 5 + 1 Ta cd log32 = = -. log75 = log72.1og23.1og35 = cab; log2 3 a i n 1 1 1 log3 7 = = = — log7 3 log7 2.1og2 3 ca 2 1 Vay 2ac + l log140 63 = 2 , 1 2c + cab + 1 abc + 2c + 1 h D H a ca Vi du 19: Trong dieu kien cd nghTa, chung minh: a) a!og<b = blog= - b) JSIaJL = 1 + ioga b l o gab x , 1 1 1 1 n(n + l) c) + + + ... + = logab loga 2 b loga 3 b loga „ b 21ogab Giai a) alogcb = blogba'°gcb = blogc b logb a =blogc" b) 1 1 ° ga X = l°gaX = loga ab = loga a + loga b = 1 + loga b loga b x log a x loga ab c) VT= + + + ...+ logab logab logab logab . - , ~ , , s 1 n(n +1) = (1 + 2 + 3 + ... + n) = logab 21ogab Vi du 20: Trong dieu kien cd nghia, chung minh: a) Ndu a2 + b2 = 7ab thi log7 -a -^ = — (log7a + logyb) 3 2 b) Neu logi218 = a, log2454 = b, thi ab + 5(a - b) = 1. c) Ndu a2 + c2 = b2 thi logb+ca + logb_ca = 21ogb+ca.logb_ca. -BDHSG DSGT12I1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
d) Neu a, b, c lap cap sd nhan thi log a d -log b d = log a d logb d - logc d logc d Giai a) a2 + b2 = 7ab => (a + b) 2 = 9ab => = Vab => dpcm. 3 , , - i ,« - l»g2 2.32 l + 21og2 3 2a- 1 b) a = log1 2 18 = —-— = — => log, 3 = log1 2 22 .3 2 + log2 3 S 2 2 - a . c< | _ log2 54 l + 31og23 _ 3b- l b = log, 4 54 = = — => log, 3 = — log2 24 3 + log2 3 3 - b Do dd ^—- = ^ ^> 6a - 2ab - 3 + b = 6b - 3ab - 2 + a 2 - a 3- b => ab + 5(a - b) = 1. c) Theo gia thiet: a2 = (b - c)(b + c). Xet a = 1 (dung). Xet a * 1 thi loga(b - c) + loga(b + c) = 2 => —-—+—-— =2 log w a log^a nen logb+c a + logb_ca = 21ogb+ca.log b_c a ' c 1 1 l0g d I d) Ta cd logad - loghd = logd a logd b (logd a)(logd b) log 1 1 Tuong tu: logbd - logcd = logd b loga c (logd b)Gogd c) Vi a, b, c lap thanh cap sd nhan ndn - = - => logd (- ] = logd (- b a {bj {aj Dodo lQ ga d ~ logb d = lQ g d c = l°ga d logb d - logc d logd a logc d V i du 21: Mdt nguoi gui 15 tridu ddng vao ngan hang theo the thuc lai kep ki han 1 nam vdi lai suat 7,56% mdt nam. Gia sir lai suat khdng thay ddi, hdi sd tien nguoi do thu duoc (ca vdn lan lai) sau 5 nam la bao nhieu tridu ddng? Giai Ap dung cdng thuc tinh lai kep: C = A(l + r) N nen sau 5 nam ngudi gui thu duoc mdt sd tien ca vdn lan lai la: C = 15(1 + 0,0756)5 ~ 21,59 (tridu ddng). V i du 22: Mdt ngudi gui 15 trieu ddng vao ngan hang theo thd thuc lai kep ki han mdt quy vdi lai suat 1,65% mdt quy. Hdi sau bao lau nguoi do cd duoc it nhat 20 trieu ddng ca vdn lan lai tir sd vdn ban dau? Giai Sd tidn ca vdn lan lai nguoi giri se cd sau n quy la: 198 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com