S = 15(1 + 0,0165)n = 15.1,0165" (trieu ddng). nen logS = logl5 + nlogl,0165=>n= logS" log15 logl,0165 De cd duoc sd tidn 20 trieu ddng thi phai sau mdt thoi gian la: log 20-lo g 15 n = — f - — * 18 (quy) = 4 nam 6 thang. logl,0165 Vi du 23: Bidt rang nam 2001, dan sd Viet Nam la 78685800 va ti Id tang dan sd nam dd la 1,7% va su tang dan sd duoc udc tinh theo cdng thuc tang trudng mu. Hdi cu tang dan sd vdi ti le nhu vay thi ddn nam nao dan sd nude ta d muc 100 tridu ngudi? Giai Theo bai ra, ta cd: S = Ae N r => 100 = 78,6858.e0 017N , do do: InlOO = ln(78,6858.e°' 017N ) => N = ^10 0 - ln78,6858 ^ ^ v ' 0,017 Vay den nam 2015 dan sd nude ta se d muc 100 tridu ngudi. Vi du 24: So 22012 ; 2 139826 9 - 1 khi vidt trong he thap phan thi cd bao nhieu chu sd? Giai Lay gia tri gan dung cua log2 la 0,3010 thi duoc: [2012. log2] + 1 = [2012.0,3010] + 1 = [605,692] + 1 = 606 Vay sd 2 201 2 cd 606 chu so. Va [1398269.1og2] + 1 = [420920, 911] + 1 = 420921 Vay sd 2 13982G 9 - 1 cd 420921 chu sd. ,13 Vi du 25: Trong khai trien nhi thuc P(x) x%/x x > 0 13 a) Tim he sd cua x Sd hang tdng quat cua P(x) = b) Tim sd hang khdng chua x. Giai 2 V 3 _2 X 3 + X\fx la: L k-H 1 '13 ( J_\ • 3 a) He sd cua x 1 3 ung vdi 13-k fxVx ) =Cj 3 .x 13k-52 13k-52 16 13 ok = 101a: T u = C\Q 3 = 286. b) Sd hang khdng chuax ung vdi 13k - 52 = 0 <x>k = 4 la T5 = C*3 =715 Vi du 26:a) Trong khai trien nhi thuc hang chua a va b cd sd mu bang nhau. a 31—Vb + VVb j tim he sd cua sd -BDHSG DSGT12/1- 199 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
b) Trong khai trien nhi thiic 200. Tim x? x lgx + l + , bidt sb hang thu tu bang a) Ta co a | b 3|-^= + -x21 'Vb VVbJ Giai ' i M a 3 b3 + • 1 b6 a 6 J 21 k=0 ^ n 2 1 - k a 3 ( i\ b2 21 21-k k k 21-k 21 42-3k 4k-21 = X c « a 3 ^ b * ~ 6 = Z c « a 6 b 6 k=0 k=0 Sb mu cua a va b bang nhau <=> 42 - 3k = 4k - 21 <=> 7k = 63 <=> k = 9. Vay he so cua so hang chua a va b co so mu bang nhau trong khai trien ,9 21! la: C 21 9!12! = 293 930. b) DK: x > 0, x * — Tacb: 10 x lgx + l + 1 2 ^ x + x 12 6-k = £C*x 2fex+1) . x 1 2 k=0 So hang thu 4 ling vbi k = 3, theo gia thiet bang 200 nen: 3 , 1 7+lgx C3x2agx+1) 4 = 20 Q x41gx+4 = 1 Q '+lg * j j 6 41gX + 4 <=>lg2 x + 31gx- 4 = 0<=> "lgx = l o lg x = -4 x = 10 „ (Chon) x = 10~4 DANG 2: CAC HAM S 6 MU, LU? THUA, LOGARIT - Tap xac dinh: y = a x ( a > 0 ) : D = R y = logax (a > 0, a * 1): D = (0; +oo) y = xn (n e N*): D = R y = x m ( m e Z \ N *): D = R \ {0} y = x a ( a € R\Z): D = (0; +oo). - Gidi han: lim ax = +oo (vdi a > 1); lim logax = +oo (vdi a > 1) l im ax = 0 (vdi a > 1); lim logax = -co (vdi a > 1) lim ax = +oo(vdi0<a< 1); lim loggX =-co(vdi0 <a< 1) 200 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
l im a x = 0 (voi 0 < a < 1); lim logax = +oo (vai 0 < a < 1) l im X-»+oO X + — = e; lim . eK -l ln(l + x ) 1 ; lim x->0 X = 1 x->0 x - Dao ham trong dieu kien xac dinh: (e x )' = ex , (e u y = eu .u', (a x )' = ax lna, (a u )' = au .lna.u' (lnx) 1 = -;(ln|x|)' = 1 ;(loga x)' =-L _ ; (ln|u|)' =H ! x x xln a - Dd thi va quan he ddi xung: y u - Tinh don dieu tren tap xac dinh: Khi a > 1: ham sd y = ax , y = logax ddng bidn tren D. Khi 0 < a < 1: Ham so y = ax , y = logax nghich bien tren D. Vi du 1: Tim tap xac dinh cua cac ham sd sau:" b) y = (x2 a - 4x + 3) ^ ) y = (x2 - 4x + 3V~5 c) y = (x3 - 3x2 + 2x) 3 d)y = 3 / x 3 -3x 2 + 2 x Giai a) Ham sd xac dinh khi: x 2 - 4x + 3 * 0 <=> x *1 va v ^ 3 ^ Vay D ~ R\ {1;3} . ill .-BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
b) Ham so xac dinh khi: x 2 - 4 x + 3>0ox< l hoac x > 3. VayD = (-oo; l)u(3;+» ) c) Ham so xac dinh khi: x3 - 3x2 + 2x > 0 <=> x(x2 -3 x + 2)>0»0<x< l hoac x > 2. VayD = (0; l)u(2;+oo ) d) Ham sd xac dinh vdi moi x ndn D = R. V i du 2: Tim tap xac dinh cua cac ham sd sau: a) y = lg(x2 ' - 9) b) y = lg(x + 3) + lg(x - 3) c) y = lo g l (4x-l)- l d)y = Jlog^(x 2 -V5. x + 2) Giai a) DK:x 2 -9>0<=>x<-3hoacx>3 . Vay D = (-oo; -3) u (3; +oo) b) DK: x + 3> 0 <=> x>- 3 [x-3> 0 x > 3 f4x-l> 0 c) DK: ]i ogl (4x-l)> l ^ ox>3 . Vay D = (3;+oo) f4x-l >0 1 (ham nghich bidn) I 4x - 1 < — » - < x < - . Vay D = (- ; - ] 4 3 4 3 J d) DK: l x2 - S x + 2 > 0 log^(x2 -75 x + 2)>0 ox2 ->/5x + l>0-»x< V5-1 VayD = 2 Vs + i <=> x2 - %/5x + 2 > 1 (ham nghich bidn) hoac x > ;+co V i du 3: Tim tap xac dinh cua cac ham sd sau: a)y 5-3 x 3X - 1 c) y = ^log x + log(x + 2) b)y = ^4X +2 X -12 d) y = A /41og2 x-log2 x-3+v'x2 -7x + 6 Giai a) DK: ^—5- >0o(5 - 3X )(3X - 1) > 0, 3X - 1 * 0 3X - 1 ol<3 x <5»0<x < log3 5. Vay D = (0; log3 5]. b) DK: 4X + 2X - 12 > 0 <=> 2X < -4 hoac 2X > 3. o 2X > 3 o x = log2 3. Vay D = [log2 3; +oo). c) DK: logx + log(x + 2) > 0 202 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
[log[x(x + 2)]>log l Jx 2 +2x-l> 0 x > 0 x >0 rx<-l-v^hayx>- l + V2 ^ x > _ 1 + ^ [x >0 Vay D = [-1 + V2 ; +00) x >0 d) DK: x2 -7x + 6>0 4 log2 x - log2 x - 3 > 0 [0<x<l hay x>6 [2 <x<8 Vi du 4: Chung minh cac gidi han: x > 0 x < 1 hay x > 6 1 < log, x < 3 a) lim — = In a x-»0 x c) lim | 1 + — o 6 < x < 8 Vay D = [6; 8]. loga(l + x) 1 b) lim- .. \Jl + ax - 1 a d) lim = — x->0 x n Giai e xlna -1 lim . In a = In a a) lim - = lim x-»o x x->-0 x x->-0 xln a b) lim1 ° g - (1 + x)=limlog.e^-+x) - x->0 x x _ > 0 X In a c) lim 1 + - I X V aj .. \/ l + ax - 1 .. 1 + ax + l a a) lim = iim — , — , = — x^o x x(^(l + ax) n _ 1 + ^/(l + ax) n " 2 +... +1) n Vi du 5: Tim cac gidi han sau: 2 „3x+2 e - e a) hm x->0 x b) lim „2x „5x e - e c) lim 2X +5" - 2 0 3X + 5X - 2 d) lim x->0 x sin4x x^o e 2 x -7 X Giai a) lim x->0 X e 2 _ e 3x+ 2 e 2 (1 — e 3 x ) = lim = -3e 2 . lim e 3 x - l x->0 3 X = -3ez -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
e 2 x -e 5 x b) lim = lim x->0 x x_> 0 t 2x e •1 e 5 x - n = 2- 5 . .. 2 x + 5 x - 2 c) lim— : = lim f - l 5 X - 1 x + x In2 + ln5_lnl0 ° 3 x + 5 x - 2 »o3" - l 5 X - 1 In3 + ln5 In 15 d) lim £an4 x = 4. x x sin4x 4x x->0 7X x->o e 3 x - l 7 x _ i 3-I n 7 X X Vi du 6: Tim cac gidi han sau: , .. ln(l + 3x) a) hm x->0 x b) lim ln(l + 3x2 ) c) lim 4x log3 (l + 5x) d) lim x-"0 l-cos2 x 6X -3X x ->o ln(l + 6x) - ln(l + 3x) Giai a) limln(1 + 3x)=3.1imln(1 + 3x) = 3 i->o x x->° 3x .. ln(l + 3x2 ) .. ln(l + 3x2 ) b) hm = lim - x ->o l-cos2x x ^> 2 sin x ili m 2 x ^o 31n(l + 3x2 ) c) lim 4x 3x2 . limsm x 3 2 5x = -ln 3 x ^o log3 (l + 5x) 51og3 e x ^oln(l + 5x) 5 d) hm 6 X -3 X r 6 x - l 3 x - l l = km x ^o ln(l + 6x)-ln(l + 3x) x -*> = (In6-ln3): (6 - 3) = -In 2. 3 Vi du 7: Tim cac gidi han sau: ln(l+6x) ln(l + 3x)N a) lim 1 + 1 x - 3 b) lim x + 3 X-V-HB I X+1 , ,. V l + 2x - V l + x c) lim x ->-° tanx d) lim Vi+^ 2 " x ^o ln(l + x 2 ) Giai a) lim 1 + — X "M<?^ x - 3 204 : lim X—>-KO 1+- x - 3 x-3 = e =e -BDHSG DSGTUI Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
b) lim —+ — j = Hm X-»+°o I X + 1 J x-woc 1 + - X +1 J * = lim 1 + - 1 x + 1 2 ; . ,. 3 yi+2x-VTT ^ c) lim = hm *->o tan x "->o vT+2x- l 4 /T+x"-l N | sinx 2 1 5 xcosx 3 4 12 -2x' d) Um -vT7 7 lim ln(l + x ) e-2x" - 1 vT+1^ = lim x->0 -2x' -2x2 va+x'f+vr+^+ i Vi du 8: Tim dao ham cua ham sd sau: a) y = (x - l) 2 x .e 2 x 2 X - 2 _ x ln(l + x 2 ) ln(l + x 2 ) 7 c) y ; 2 X + 2" x b)y = x 2 V e 4 x + l d) y = x5 - 5X + xx Giai a ) y> = e2x + ( x _ i).2e 2 x = (2x - l)e 2 x , . ri— 2x2 e 4x 2x[(x + l)e4x+l] = 2xVe +1 + : = —-—. =± Ve 4 x + 1 Ve 4 x + 1 (2X In 2 + 2~x In 2)(2X + 2"x ) - (2X - 2"x )(2X In 2 - 2~x In 2) b) y c) y' = (2x +2- x ) 2 ( 2 x +2- x ) 2 -(2 x -2- x ) 2 , 2 o 41n2 2 — m 2 = (2 X +2" x ) 2 (2 X +2" x ) 2 d) Tacdy = x 5 - 5 x + xx = x5 - 5X + e xln x nen y' = 5x4 - 5x ln5 + e xlnx (lnx + 1) = 5x4 - 5x ln5 + xx (lnx + 1). Vi du 9: Tim dao ham cua cac ham sd sau: a) y = (3x - 2)ln2 x b) Vx2 + llog 3 x2 c) y = ln(x + %/x^ + a 2 ) e) y = log^(-x2 +5x + 6) d)y ln(x2 +1) a) y' = 31n2 x + 2(3x-2)ln x f) y = cosx.e 2tan x Giai b)y=x ; 0g3x2 + 2 ^ln3 Vx2 +1 x -BDHSG DSGT12/1- 205 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
1 + c) y' = e) y' = 1 d)y' = x + Vx^+ V Vx 2 + a 2 * + 1 -2x + 5 _ -4x + 10 (-x2 +5x + 6)lnV3 ~ (-x2 +5x + 6)ln3 2 ln(x2 +1) •2 9 f) y' = -sinx.e .( 2 cosx I cosx - sin x V i du 10: Tim dao ham cua cac ham so sau: b)y = d)y = a)y = (2x+l) n -tane x b) y = Vin 3 5x c)y 1 + x 3 1-x 3 X b Giai a / \b a vdi a > 0, b > 0 a) y' = 2rc(2x + l)"" 1 - (1 + tanV)e* , _ Qn3 5xV 3In2 5x = 3 Y " 5#n3 5x)4 " 5vV2 5x 5vV 5x 1 + x3 , , , u' , c) Dat u = j thi y' = —->= va u' = 6x^ l-x J ,3r 3Vu2 3 \2 (1-x 3 ) nen y d) y' = u'Vu 2x2 J l + x3u 1 - x 6 U - x b 'a^ b t \ a _ 1 fa^ b a x l fa^ X \ - - + - b IbJ v b j a x b-l \a / \b , x i a a- b b J { x ) x V i du 11: Chung minh: a) Ndu y = hi—!— thi xy' + 1 = ey 1 + x b)Neuy= — thi y(4) + 4y = 0. ex c) Neu y = e4x + 2e-x thi: y'" - 13y' - 12y = 0 Giai 1 -x 1 1 y a) y' = suy ra xy' + 1 = +1 = 7 - e ' y x + 1 x + 1 x + 1 b) Ta cd y = cosx ~x .cosx => y' = e x (-sinx - cosx) 206 y" = e~x .(2sinx), y'" = 2e~x (cosx - sinx), y( 4 ) - -4e x.cosx Do ddy(4) = -4y=>y(4) + 4y = 0. -BDHSG DSGT1Z/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
c) y' = 4e 4 x - 2e'\ y" = 16e4x + 2e'x , y'" = 64e 4 x - 2e~x nen: y"' - 13y' - 12y = (64e 4 x - 2e"x ) - 13(4e4x - 2e"x ) - 12(e4x + 2e"x ) = 0 V i du 12: Tim dao ham cap n cua ham so a)y = 5k x b)y = ln(x-5 ) c) y = ln(6x2 - x - 1) Giai a) y' = (kln5).5k x , y" = (kln5) 2 .5k x Ta chung minh quy nap: y( n > = (kln5) n .5k x v - c . 1 - 1 ... 1-2 b) Voix * 5: y = ; x - 5 y"= - —; y Ta chung minh quy nap: y(n) c) Vdi x < -— hoac x > — 3 ' 2 (x-5) 2 ' J (x-5) 3 (-l)-i(n-l) ! (x-5) n y = ln((2x - l)(3x + 1)) = In | 2x - 11 + In | 3x + 1 y i i r + • 2x- l 3x + l Ta chung minh quy nap 1 I ax + b (m) (-l) m m!a m (ax + b) m + 1 Suy ra y<"> - (-^(n-l) ^ + (-D^ n - 1)! ^ (2x-l) n (3x + l) n Vi du 13: Tim khoang don dieu va cue tri ham sd: e „2 „-x a) y x c) y = ln(x2 - 1) b) y = x2 .e" d) y = x - ln(l + x) Giai a) D = R\{0},y l =%^,y ' = 0ox=l , BBT X —00 0 1 +00 y' - - 0 + y +00, +00 * e +00 Vay ham sd nghich bidn trong cac khoang (-oo; 0) va (0; 1), ddng bien tren khoang (1; +oo), dat CT(1; e) b) D = R, y' = (2x - x2 )e x , y' = 0 o x = 0 hoac x = 2. BBT y -BDHSG DSGIT2/1 0 2 +00 0 0 +00. r4e: 2U7 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Vay ham sd ddng bien trong khoang (0; 2). nghich bien trong cac khoang (-00; 0) va (2; +00), dat CD(2; 4e"\ CT(0; 0). c) D = (-00; -1) u (1; +00), y' = -p - x - 1 Khi x < - 1 thi y' < 0 nen ham so nghich bien tren (-co; -1). Khi x > 1 thi y' > 0 nen ham sd ddng bien tren (1; +00). Ham sd khdng co cue tri. d) D = (-1; +<»), y' = !- • 1 -. y' = 0 <=> x = 0. 1+x 1+x y' > 0, Vx e (0; +00) nen ham sd ddng bidn tren (0; +00). y' < 0, Vx e (-1 ; 0) nen ham sd nghich bien tren (-1 ; 0). Ta cd y'' = —-— r > 0 nen dat cue tieu tai x = 0, y d = 0 (1 + x) 2 ^ • ' V i du 14: Tim khoang dong bien, nghich bien cua ham sd: b)y a)y-| f c) y = log2 x a) V i co sd — > 1 nen ham sd ddng bidn tren D = R. 3 .V2 + V3 d) y = logax vdi a Giai 3(V3 - 72) b) Vi co sd V2+V3 1,4 + 1,7 < 1 nen ham sd nghich bien trdn D = R. c) V i co sd — < 1 nen ham sd nghich bien tren D = (0; +00). d) Vi co sd S + V2 > 1 nen ham sd ddng bidn tren D = (0; +00). 3(V3-V2) 3 V i du 15: Ve dd thi ham sd y = f(x) = 2X Suy ra dd thi cac ham sd y = 2X - 1, y = 4.2x,y = -2 x , y = A x , y = 2 l x l Giai y = f(x) = 2X , D = R. lim y = +00, lim y = 0 => TCN: y = 0 (khi x —» -00) y' = 2\ ln2 > 0, Vx nen ham so dong bien tren D = R. BBT X —CO +00 y' + y 0 *- +00 208 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Cho x = 0 => y = 1 X= 1 =>y = 2 1 x = - l y = 2 Ta co: y = 2 X - 1 = f(x) - 1: Tinh tien xuong duoi 1 don vi y = 4.2X = 2 X + 2 = f(x + 2): Tinh tii n sang trai 2 don vi. y = -2 X = -f(x): Lay doi xung qua Ox. 2"x f(-x): Lay doi xung qua Oy. l o y = 2 i x | =f(|x| ) ham sd chin, khi x > 0 thi y = f(x) nen lay phan nay va lay ddi xung cua nd qua Oy. Vi du 16: Ve dd thi ham sd y = f(x) = log2X. Suy ra cac dd thi ham sd y = log22x, y = lg2 (x - 3), y = log2 (-x), y = log^ x, y = log2 1 x |. 2 Giai y = f(x) = log2 x, D = (0; +oo) lim y = +oo, lim y = -oo => TCD: x = 0 (khi x -» 0 + ). x->+« x->0* 1 > 0, Vx > 0 nen ham sd ddng bien tren (0; +oo). xln 2 BBT X —OO +00 y' + y —00 " Chox 1 -1 X = l :=> y = 0, x = 2 => y = 1. Ta cd: y = log2 2x = f(x) + 1: Tinh tien len tren 1 don vi y = log2 (x - 3) = f(x - 3): Tinh tidn sang phai 3 don vi. y = log2 (-x) = f(-x): Lay ddi xung qua Oy. y = logj x = -f(x): Lay ddi xung qua Ox y = log2 1 x | = f( | x |) la ham sd chan. khi x > 0 thi y = f(x) nen lay phan nay va lay ddi ximg cua nd qua Oy. Vi du 17: Khao sat su bien thien va ve dd thi cua cac ham sd sau: i i a) y. b)y = x c)y Giai a) D = R , ham sd le, y' = -3x~ = —— < 0. Vx ^ 0 nen ham sd ludn nghich x -BDHSG DSGT12/1- 209 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
bien tren cac khoang (-00; 0) va (0; +00). Ta co lim y = lim y = 0 , lim y = +00, lim y = -co nen tiem can ngang la x->-=o x->+oo x->0* x-»0 y/ true hoanh, tiem can clung la true tung. BBT X -co 0 +°c y1 - - y —00 0 o Do thi cua ham so co tam doi xung la gdc toa dd. 1 -- b) D = (0; +00), y' = — . x 2 < 0, Vx > 0 nen ham sd nghich bien tren (0; 2 +00). Ta cd lim y = +00, lim y = 0 ndn tiem can dung la true tung, tiem can ngang la true hoanh. BBT: X 0 +00 y' - y +00 Cho x = 1 => y = 1. c) D = (0; +00), y' 37x2 > 0, Vx > 0 nen ham sd ddng bien tren (0; +00). Ta cd lim y = 0 , lim y = +co => khdng cd tiem can X->0+ X->+=o BBT: X 0 +00 y' + y + +00 0 — Cho x y=l. x = 8=>y = 2 L 2 1 1 1 O 1 8 x a) y = ax va y 1 v a ddi xung vdi nhau qua true tung. b) y = logax va y = logl x ddi xung nhau qua true hoanh. a c) y = a x va y = logax ddi xung vdi nhau qua phan giac 1. 210 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Giai Goi M(x 0 ; y0 ) la mot diem bat ki. Khi do diem doi xung vdi M qua lan luot: a) True tung la M'(-x 0 ; y0 ). Ta cd: (1 Vx ° M 6 (G,)oy 0 = a -oy 0 = - o M ' e (G2 ) Dieu do chung td (Gi) va (G2) doi xung vdi nhau qua true tung. b) True hoanh la M"(x0 ; -y 0 ). Ta cd: M G (G.) o Yo = logax0 » -y 0 = logj x0 » M" e (G2) a c) Phan giac 1 la M'"(y0 ; x0 ). Ta cd: M e (Gd o y0 = a x ° o x0 = log!1 y0 o M" ' G (G2 ). 4 X Vi du 19: Cho ham so f(x) = kf ' 1 Tinh tdng: S = f i De y neu a + b = 1 thi: 4a 4b 4 X +2 2 n n - l 4 a + 2 4"+ 2 + ... + f Giai 4a (4b +2) + 4b (4a +2) (4a + 2)(4b + 2) i+b + 2.4a +4 +2.4b 2.4 a +2.4 b +8 Ta cd: S = f n-l 4 a+b +2.4 a +2.4 b + 4 2.4 a +2.4 b +; - l + f hay S = f n + f - • • ..ff^ 1 I n n-2 ^ + ... + f f I n t u S _n- l 2 ap dung thi 2S = n - 1 => S Vi du 20: Goi (C) la do thi cua ham sd y = lnx va d la mdt tiep tuyen bat ki ciia (C). Chung minh rang tren khoang (0; +oo), (C) nam d phia dudi cua dudng thang d. Giai Goi Xo la hoanh dd cua diem M tuy y thudc (C). Tiep tuyen d cua (C) tai M cd phuong trinh y = — (x - XQ) + lnxo. Khang dinh can chung minh tuong duong vdi bat dang thuc sau: Vx e (0; +oo). 1 -(x-x 0 ) + lnx 0 - In x > 0 <=> — - In — > 1 -BDHSG DSGT12/1- 211 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Xet g(t) = t - lnt voi t > 0, g'(t) = 1 - i = ^ BBT »'(t) = 0ot = 1. X —00 1 y' - 0 + y ^"~~~*- 1 Ta co g(t) > 1, Vt> 0 => dpcm. V i du 21: Cho ham so y = -x J + 3x - 2 a) Khao sat su hien thien va ve do thi (C) cua ham so da cho. b) Bien luan theo m so nghiem cua phuong trinh -x 3 + 3x - 2 = logsm. Giai a) Phan khao sat danh cho ban doc. Dd thi: b) Sd nghidm cua phuong trinh la sd giao didm ciia dd thi (C) vdi cac dudng thang y = log3m. - Neu m < 0 thi phuong trinh vd nghiem. - Neu m > 0, dua vao dd thi (C) va dudng thang y = log3tn ta co: Khi log3m <-4o0<m < — , phuong 81 trinh cd mdt nghiem. - 2 \ -1 k 1 0 -2 \ X y = log,m -4 Khi log3m = -4 o m = — . phuong 81 trinh cd hai nghidm Khi 0 > log3m > - 4 <x> 1 > m >— , phuong trinh cd ba nghiem 81 Khi log3m = 0 <=> m = 1, phuong trinh cd hai nghiem Khi log3in > 0 <=> m > 1, phuong trinh cd mdt nghiem. DANG 3: BAT DANG THUC VA GTLN, GTNN - So sanh cung co sd ciia mu: a > 0, a ^1 . Ndu a > 1 thi: a M > a N o M > N . Neu 0 < a < 1 thi: a M > a N <=> M < N . - So sanh cung luy thua ciia mu: 0 < a < b ax < bx <=> x > 0: a x > bx o x < 0. - So sanh cung co sd cua logarit: a > 0, a 1 Neu a > 1 thi: logaE > logaF o F > F > 0 Neu 0 < a < 1 thi: logaE > logaF o 0 < E < F - Su dung cac bat dang thuc co ban, danh gia dao ham cac cap cua ham so, phdi hop bien ddi tuong duong, so sanh va tinh don didu, lap bang 212 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
bien thien cua ham so ete chung minh bat dang thuc va tim GTLN, GTNN. Vi du 1: So sanh cac sd: a) 72 va 73 c) 73 + 730 va 763 b) 7l 3 va 723 d) 77 + 7l 5 va 7l 0 + 728 Giai a) Tacd(7 2 ) 6 = 23 = 8; (73) 6 = 32 = 9. Do9>8ndntacd(72)6 < (73)6 , suy ra 72 < 73 b) 7l3 = 27l3^ = 27371293 ; 123 = 2 7234 = 2 7279841 Ta cd 371293 > 279841 nen 7l 3 > 723 c) 73 + 730 > 1 + 727 =4 = 764 > 763 d) 77 + 7i5<2 + 4 = 3 + 3<7l0 + 728 Vi du 2: So sanh cac sd: 6 va 3 3_ 1 4 - b) 3600 va 54 0 0 c) va 72.21 4 d) Giai 73") 4^5 va 3 -3V2 (73p= 3 ^ va ^S- 1 ^ ! =; 3 /3- 1 - 1 ~T 34 12 Vi co sd 3 > 1 nen (VSJ «<?ls-'il1 ,600 ( 3 3) 2 00 27200 b) Ta cd: 3 va 5 400 = (5Z YW = 25200 y . y 3 600 > 5 400 c) 1 2 Vay 5 _3_ 1 _3_ 1 _3_ 5 V ; 72.21 4 =22.2l 4 =2 2+1 4 =2 7 : 72.21 d) Ta cd 1 f^ 2 ^ va 3 -3V2 ^\3V2 3 j Ta cd 372 < 275 c=> (372) 2 < (275) 2 <=> 18 < 20 (dung). N 2V5 /^\3V2 / . \4V5 1 V i co so 0 < - < 1 nen | -BDHSG DSGT12/1- ,/3 <3 -3 72 213 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
V i du 3: So sanh p va q biet: a) c) 0,25p < 2q 2 2 a) b) c) 0,25p < b) Giai — < 1 => p < q . 3 P-2q d) 1^2q 2, P-2q - I 3 > 1 : -p < - q => p > q , -<l=>p> q 4 2q-p >l=>p<2q-p=>p<q V i du 4: Hay so sanh: a)log34v log4^ c) log827 > log925 b) 3 log6 1,1 7 log6 0,99 d) log4 9 > log925 Giai 1 a) Ta cd log34 > 1 va log4- < 0, suy ra log34 > log 4 - 3 3 b) Ta cd log 6 l , l > 0 nen 3'° S r " > 3° = 1 (vi 3 > 1) va log60,99 < 0 nen 7 log6 0,99 <7 o = 1 (y i ? > {y Suy ra 3log6U >71og60,99. c) log827 > log825 > log925 d) log49 = log23 = log8 27 > log9 25. V i du 5: Khong dung bang so va may tinh, hay so sanh: a) log2 + log3 vdi log5 b) logl2 - log5 vdi log7 c) 31og2 + log3 vdi 21n5 2 • 3 d) log3 - va log3 - 5 3 2 5 Giai a) log2 + log3 = log6 > log5. 12 b) logl2 - log5 = log— = log2,4 < log7 5 c) 31og2 + log3 = log(23 3) = log24 < log25 = 21og5 < 21n5. d) Vi - < 1 va - < 1 nen log3 - > log31 = 0 5 3 5 3 4 214 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
V i - > 1 va - < 1 nen log3 - < log3 1 = 0 2 5 2 5 2 2 3 Tir do suy ra log3 - > log3 - 5 3 2 5 Vi du 6: Chung minh: a)—— + —^— >2 b) log23 > log34 log2 n l °g5 n c) logn(n + 1) > logn+i(n + 2) vdi moi so nguyen n > 1. d) am + bm < cm , niu m > 1, a + b = c vdi a > 0, b > 0. Giai a) Ta cu: —— + —!— = log.5 + log„2 = logJO > log^ = 2 l0g2 7T l0g5 71 b) log23 > log34 <=> —-— > log34 <=> log32.1og34 < 1: Dyng vd log3 2 j\og3 2,log34 < - aog32 + log34) = i log3(2.4) < i log39 = Z Z Zi c) A = logn(n + 1) = lognn(l + -) = 1 + logn(l + -) n n B = logn+1(n + 2) = logn+1(n + 1) (1 + ) n + l 1 + logn + 1 (1 + n + l ) Ta cd 1 + — > 1 + n ^log n ( l + I)>log„(l+-!- ) n + l n n+ l va logn ( 1 + — ^-)>log n + 1 ( 1 + —^- ) n + l n+ l logn(l + -) > logn+, ( 1 + —?—). Do do A > B. n n + l d) Tacda m + b m < c m o •l- l Ma a + b = c, a > 0, b > 0 nen 0 < - < 1. 0 < - < 1 c c Suy ra vdi m > 1 thi Tir do ta cd: Vidu7: (a) m m - < - ; - < - (a) m 'b) m a b - + - < - - c c 1 a) Chung minh: nn > (n + l) n , Vn e N, n > 3 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
x - - |>8 4. b) Cho 4 sd x. y, z, t e (- ; 1). 4 Chung minh: Giai a) Vai n e N, n > 3, bat dang thuc tuong duong , , ,x n + l n (n + l)lnn > nln(n + 1) « > ln(n + l) In n Xet f(x) = — tren (3; +oo) thi f '(x) = lnX" 1 > 0 Do do f ddng bien In x In x tren (3; +oo) nen: n + 1 > n > 3 => f(n + 1) > f(n) => (dpcm). iV i > 0 => a < a 2 vdi moi a. 4 b) Ta cd: a — 2 Va vi — < x, y, z, t < 1 nen ham nghich bien, do do: 4 VT > lc^y2 + logyz 2 + logzt 2 + logt x2 = 2(logxy + logyz + log.t + logt x) > 8.4/logx y.logy z. logz t.logt x = 871 = 8 . Vi du 8: Chung minh cac bat dang thuc sau vdi moi x > 0. 2 2 a) ex > x + 1 b) ex > 1 + x + — c) ln(l + x) > x - — 2 2 Giai a) Xet ham sd f(x) = ex - x - 1, x > 0 thi f '(x) = ex - 1 > 0, Vx > 0 nen f ddng bien tren (0; +°o) vi f lien tuc tren [0; +oo) ndn f ddng bien tren [0; +oo): x > 0 => f(x) > f(0) = 0 ^> dpcm. b) Xet f(x) = ex - — - x - 1, x > 0 thi f '(x) = e x - x - 1. Theo cau a) thi f '(x) > 0 ndn f ddng bien tren [0; +oo). x > 0 => f(x) > f(0) = 0: dpcm. x 2 c) BDT:ln(l + x) - x + — > 0, Vx > 0 2 2 Xet f(x) = ln(l +x )-x+—,x>0 , f '(x) = ^— > 0 2 1 + x va f lien tuc tren [0; +oo) nen f ddng bien tren [0; +oo) Do do: x > 0 => f(x) > f(0) = 0 => dpcm Vi du 9: Chung minh: a) 4 s m x +2 tan x > 72 3x+ 2 , Vx e (0; - ) 216 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
b) e x > — vbi moi x x 2 -2 x + 2 Giai a) Ap dung bat dang thuc Cdsi: ^sinx 2 tan x > 2yj\^s m x 2 t a n x — -\/22s^n x+tanx+ 2 Ta can chung minh: 2 2si n x + ,anx + 2 > 2 3x+ 2 « 2sinx + tanx > 3x Xet f(x) = 2sinx + tanx - 3x, 0 < x < — f '(x) = 2cosx + — 3 > 2cos2 x + — 3>2>/2-3>0 cos x cos X nen f dong bien tren [0; —): x > 0 => f(x) > f(0) = 0 ^> dpcm 2 b) Neu x < 0 thi BDT dung. Neu x > 0, vi x2 - 2x + 2 > 0, Vx nen BDT « x2 - 2x + 2 > . Xet f(x) = x2 - 2x + 2, x > 0 f'(x) = 2x - 2, f '(x) = 0 « x = 1. Lap BBT thi minf(x) = f(l) = 1 Xetg(x) = 4,x>0,g'(x)=^^ = ^;g'(x) = 0«x=l e e e Lap BBT thi maxg(x) = g(x) = — V i minf(x) > maxg(x) => dpcm e Vi du 10: Chung minh cac bat dang thuc sau: x2 a) ex + cosx > 2 + x . Vx 2 b) ex - e"x > 21n(x + V l + x2 ) , Vx > 0. Giai a) Xet ham sd f(x) = ex + cosx - 2 - x + —. D = R. f '(x) = ex - sinx - 1 + x; f '(x) = 0 O x = 0. f "(x) = ex + 1 - cosx > 0, Vx nen f '(x) ddng bien trdn R, ta co: f '(x) < f '(0) = 0, Vx < 0; f '(x) > f '(0) = 0, Vx > 0. BBT: X —00 0 +00 f'(x) — 0 + f(x) ———- 0 ' Vay f(x) = e x + cosx - 2 - x + — > 0, Vx 2 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
b) Xet ham so f(x) = ex - e x - 21n(x + V l + x 2 ) , D = [0; +co) 2 f'(x) = ex + e- x - vi ex + e~x > 2 va x/T+1? 2 T T : f'(x) = 0o x = 0. < 2 nenf'(x)>0 , Vx > 0. + x Do do f(x) ddng bien tren [0;+co) nen f(x) > f(0) = 0 => dpcm. Vid u 11: Cho0<x < l;0<y < 1 vax*y . Chung minh rang: 1 ln-^-ln - ' y-x v 1-y 1-x J Giai Do x * y, khong giam tdng quat, gia sir y > x t >4 Xet ham sd f(t) Vay f| hay In - 4t, vdi 0 < t < 1 f(t ) = 1 -t • t(l-t) ) Vay f(t) la ham ddng bien tren (0; 1) ma y > x ndn ta cd f(y) > f(x) y 4y > In— 4x va do y - x > 0 ndn suy ra l - y i- x l y - x i y i x In— ln- i - y 1-x >4 dpcm. V i du 12: Cho a > b > 0. Chimg minh |^2a + Giai Vdi a > b > 0, bat dang thuc tuong duong: 2a + — < 2 +- r I 2a j v 2b J 4a +1 ) <(4 a +l) b <(a b + l) a o b.ln(4' + D < a.ln(4b + 1) o < ln ^ a b Xetf(x)= ln( 1 + 4 ' ) ,x> 0 f'(x) 4 X In 4 1 + 4 X .x-ln( l + 4x ) - (4X . In4 X - (1 + 4 X ). ln(l + 4X )) < 0 nen f nghich bien: a > b > 0 => f(a) < f(b) => dpcm V i du 13: Cho cac sd nguyen n (n > 2) va hai sd thuc khdng am x, y. Chung minh: v/xn + y n > n+1 / x n + 1 + y n+l Giai Vdi x = 0 hoac y = 0, bat dang thuc dung. V d i xy > 0, bit dang thuc can chung minh tuong duong vdi 218 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
1 + ( \ n > n+l 1 + 1^ ^n+1 , n Xet f(t) = V , , voi t e (0; +<»). yj n + 7l+t n + 1 BBT \ — t n_i a-t ) ) n+l /(i+t-r^i+f)" - X 0 J +00 f'(t) 0 + 0 f(t) 1 1 ; f'(t) = 0 <=> t = 1. Suy ra f(t) > 1 vai moi t e (0; +oo) => dpcm. Vi du 14: Cho a. b>0va a + b = l . Chung minh bat dang thirc eax+b y < a.ex + b.ey , Vx, Vy. Giai Ta cd b = 1 - a do do 0 < a < 1 nen BDT: e ax+(l-a)y < & &x + (\ - a )e Y o ey ea(x-y ) < ey + a(ex " y - 1) ea(x_y ) - a.ex ^' + a - 1 < 0. Xet f(t) = e* - a.e' + a - 1, t e R. f '(t) = afe* - el ), f'(t) o t = fl. BBT X -00 0 +C 0 f + 0 f 0 Suy ra f(t) < 0, Vt => dpcm. Vi du 15: Cho p > 1, q > 1 thoa p + q = pq va a, b > 0 a p Chung minh ab < — + — P Q Giai a p bq Xet ham so f(a) = — + ab vdi a > 0. P q i f - ( a ) = a H _ b, f'(a) = 0 o a p _ 1 = b ^> a = b^ 1 Ma p + q = pq => (p -l)(q-l ) = 1 nen a = b q ' ' Lap BBT thi min f = f(bq "') = 0 => dpcm. Vi du 16: Cho a, b, c > 0. Chimg minh a + b+c a) aa .b\cc > a b .bc .ca b) (abc)^ " <a\b V c) a b + ba > 1 d) (a + b) c + (b + c) a + (c a) b > 2 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Giai a) Gia su a = max{a, b , c}. Xet a > b > c: BDToa a " b bb ~c >ca ~c V i a > b > c > 0 nen aa_b.bb ~c > ca_b .bb " c = ca _ c Xet a > c > b: BDT <=> aa ' b > bc " b .ca " c V i a > c > b > 0 nen b c _ b ca " c < ac " b .aa " c = aa _ b a+b+c b) BDT <=> log(abc)^ " < log(aa .bb .cc ) c=> (a + b + c)log(abc) < 3(logaa + logbb + logcc ) <=> (a + b + c)(loga + logb + logc) < 3(aloga + blogb + clogc) <=> (a - b)(loga - logb) + (b - c)(logb - logc) + (c - a)(logc - loga) > 0. BDT nay dung vi ca so 10 > 1 nen x > y > 0 => logx > logy hoac 0 < x < y => logx < logy nen (x - y) (logx - logy) > 0, Vx > 0, Vy > 0. c) Neu a > 1 hoac b > 1 thi ab + ba > 1 Neu 0 < a , b < 1. Xet f(x) = (1+x)" -1-ocx, x>0,0<a<l . ( f '(x) = a(l + x) a _ 1 - a = a (1 + x) 1 1 1-a < 0 . nen x > 0 => f(x) < f(0) = 0 ^> (1 + x) a < 1 + ax (*) k A 1 nb i a+ 1 Ap dung a= ,x>0=> a > 1 + x 1 + xb a + b + ab Tuong tu: ba >^^= b + 1 =>ab +ba >l 1 + ya a + b + ab d) Trong 3 so a + b, b + c, c + a neu cd mot sd, chang han a + b > 1 thi (a + b) c > 1 va (b + c) a + (c + a) b > ba + a b > 1 suy ra dpcm. Con neu ca 3 sd do be hon 1 thi dung bat dang thuc (*). V i du 17: Tim gia tri ldn nhat va gia tri nhd nhat cua ham sd a) f(x) = x - e 2 x ndn doan [-1 ; 0]. b)f(x) = 3 | x | tren doan [-1 ; 1]. c) f(x) = x - lnx + 3 tren khoang (0; +oo). d) f(x) = ln(x2 + x - 2) trdn doan [3; 6]. Giai a) Tacd:f'(x) = 1 - 2e2x , f'(x) = 0ox-lnV 2 e (-1; 0). f(-l) = -1-e"2 , f(-lnV2 ) = -1+^n2 , f(0) = -1. So sanh thi maxf(x) = f(-lnV2) = -lnV2--, min f(x) = f(-1) = -1 -e"2 xe[-l;0] 2 xe[-l;0] b) f(x) la ham sd chan va lien tuc tren doan [-1 ; 1]. Xet 0 < x < 1 thi f(x) = 3X => f '(x) = 3X ln3 > 0 nen f ddng bidn tren [0; 1 j . Vay min f(x) = f(0) = 1 ; max f(x) = f(±l) = 3 xe[-l,l] xs[-l,l] 220 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
c) f(x ) = l- i = — ,f'(x ) = 0o x = 0. X X Lap BBT thi min f (x) = f (1) = 4, khong co gia tri Ion nhat. xs(0;+co) ° ° d) Taed f'(x): 2x + l nen f'(x) > 0, Vx e [3; 6] do do tren doan [3; 6] x" + x- 2 ham sd f(x) ddng bidn. Vay mm f(x) = f(3) = In 10 ; maxf(6) = In 40 XE3,C xe 3,6 V j du 18: Tim GTLN, GTNN cua y = 21 sin x 1 + 21 cos x 1 Giai Datt = | sinx |, 0 < t < 1, thi y = f(t) = 2l + 2^ , 0 < t < 1. f'(t) = 2 tln2 + 2 V l - t 2 U - t .In 2 =t.ln2 2l Xet g(u) = — , 0 < u < 1 thi g '(u) = 2 U uln2- l u i r V i 0 < u < 1, 0 < ln2 < 1 nen g '(u) < 0, Vu e (0; 1), do dd g(u) nghich bien tren (0; 1) ndn: f '(t) = 0 o 2l o t 4i Tac6f(0) = 3, f _y_2 , 2 , 4i 2.2 2 ,f(l ) = 3. Vay max y = 2.2 2 , min y = 3. Vi du 19: Tim GTLN, GTNN cua y = 4 S1I,2 T + 4C Giai Datt = 4 s m \ 1 < t < 4 thi y = f(t) = t + - . f'(t) = 1- ^ 4 t 2 - 4 t t 2 t 2 f'(t) = 0o t = ±2 . Chon t = 2. Ta cd f(l) = 5, f(2) = 4, f(4) = 5 nen max y = 5 khi sin2 x = 0 hoac sin2 x = 1, min y = 4 khi sin2 x = — , " 2 Vi du 20: Tim m de phuong trinh sau co nghidm thuc: 3%/x-l + mVx + 1 = 27x2 - 1 Giai Dieu kidn x > 1 dat 4/2E—1 ^ t = JiLJ : = A\ - —?— nen 0 < t < 1. x + 1 x + 1 x + 1 Phuong trinh da cho: 3./ — + m = 2}i—-— hay la -3t 2 + 2t = m. Vx+ 1 Vx+ 1 Xet ham sd f(t) = -3t 2 + 2t, 0 < t < 1. WHSG DSGT12/1- 221 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Ta co f '(t) = -6t + 2, f'(t) = 0 o t BBT t 0 1/3 1 f + 0 f -1/3^ ^ Vay phucmg trinh da cho co nghiem khi - 1 < m < — 3 V i du 21: Tim m de phuong trinh sau cd dung 2 nghiem: 4 /2x + 72x + 276- x + 276 - x = m Giai Xet f(x) = V2x + 72x + 2^6 - x + 2x/6 - x , 0<x < 6. Vdi 0 < x < 6, ta cd: f '(x) = - 2 (^(2x7 7(6-x) 3 J W2 x 76 ^ f(2) = 0, x > 2 => f '(x) > 0, x < 2 => f '(x) < 0 BBT: X 0 2 6 f 0 f Ta cd f(0) = 276 + 276 , f(2) = 6 + 3 72 , f(6) = 7l 2 + 273 Vay dieu kidn cd 2 nghidm phan biet 276 + 276 < m < 3 72 +6. C. BA I LUYE N TAP Bai l.a) Tinh 21 ; (4,72)°; (-2) 2 ; (3) 4 ; (-4)" 3 b) Tinh 2:4" 2 +(3" 2 ) 3 .(-)" 3 9 5-3 .252 +(0,7)°.(-)- 2 DS: b) 11/3 Bai 2. Viet dudi dang luy thua vdi sd mu huu ti cac bieu thuc sau day: a) 7x4 .7x (x>0 ) Bai 3. True can d mau sd cua cac bieu thuc sau: a) 1 va b (a>0, b>0 ) b) 1 73 + 72 222 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
c)— - d) 4-V1T DS: c) 4 + Vl l Bai 4. Gian udc cac bieu thuc sau: a) V(a - 5) 4 b) 7x 8 ( x + l) 4 vdi x < - 1 c) (4 - x)J—— ne'u x > 4 d) (5 - a) I —- neu 0 < a < 5 x - 4 V25- a Bai 5. Tinh: -3 3 a) (2a 4 +3a 4 ) 2 (a > 0) -1 2 -2 4 2 -1 b)(a 5 +a 5 )(a 5 + a5 )(a^-aY ) (a > 0) c) (Vx - Vx + l)(Vx + Vx" + l)(x - Vx + 1), X > 0 d) Vx2 -2 x + l + Vx2 + 2x + l DS: c) x2 + x +1 Bai 6. Chung minh: Va 2 +Va 7 b T + Vb2 + Va^b1 = ^(Va 7 + Vb2 ") 3 H D: Binh phucmg tuong duong Bai 7. Tinh gon: a) A = ^10 + 8^2 +V9 + 4V2 b) B = V38 + 17Vs c) C = V20 + 14V2 + V20-14V2 d) D = (V25 + 4V6 - V l + 2V6)Vl-2V6 DS:c)4 d)0 Bai 8. So sanh: a) 74 va V5 b)2v a 2 72 + 3 73 c)VTT + Vl4 va Vl2+VT3 d) .— 1 ;— va 759-75 8 Bai 9. Cho a, b > 0, a + b = 2. Chung minh: ab < a".bb Bai 10. Ham sd nao ddng bien, nghich bien? a) y = Ax b)y = (-)x 3 e 3K _ 3-' Bai 11. a) Chung minh ham sd sau don dieu: y = 58 -V57 2 b) C/m ham sd y = 2t g x ddng bien trong cac khoang (--+kJt; - +krt) vdi keZ . -••''<'•>' 2 2 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Bai 12. Ve do thi cua cac ham so sau: <5 a ) y = 3x b) y = A x c) y = -3 X d) y = 3b Bai 13. Tim x biet: a)2x =16 b)3x =- c)5x =0 9 d)2x <1024 e)(-)x >27 f)5x > — 3 125 g)2 x =l- x h)2 x =x- 3 k)2 x = x + l Bai 14. Cho a > 1. Vdi gia tri nao cua x thi do thi cua y = ax nam d phia tren dudng thang y= 1; nam d phia dudi dudng thang y = 1. Cung cau hdi nhu tren, nhung dieu kien 0 < a < 1. Bai 15. Tim a > 0 trong cac trubng hop: a s <v? ; a 2 > a 3 ; (2a 2 - a + l) x2+ 1 < (a 2 + 3) x2+1 ,Vx Bai 16. Tim: 1 *x2 + 4x a) GTNN: y = 2 X + 2" x b) GTLN: y = (-) 3 c) GTNN, LN: y = 3 1+2cos x d) GTNN, LN: y = 5s1" 2 x + 5cos2 x DS:a)2 b) 81 c)l/3va2 7 Bai 17. Tim x theo y biet: a)y = x5 -3 b) y = lzi.(x*-l) x + 1 c ) y = x 2 - 2 x + 2 d) y = x 2 - 3 x tren [|;+oo] Bai 18. Ve dd thi: a) y = log3 x b) y = log_ x 3 c) y = |log3 x| d) y = log3 |x| Bai 19. Tim mien xac dinh ham sd: a) y = Jlog2 x b) y = log 3 ( x 2 +x-6 ) c) y = log 3 |x-2 | d)y = log 5 ( x 2 + 4 x + 4) e)y = \Kfi^ f)y^log,(x-3)-i DS: a) x > 1 c) x * 2 e) x > 1 Bai 20. Ve do thi: a) y = log2 (x + l) b) y = log 2 x 2 c) y = log 2 |x-3 | d) y = 2 log2 X e) y = log2 (x + l) + log 2 ( x - l ) f) y = log 2 ( x 2 - l ) 224 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Bai 21. Tinh: a)log216 b) log3^y c) log10(0,00001) d) log5 625 e) log81 27 f) log0 5 8 g) log^ 243 h) log^ 8 i) logl 373 Bai 22. Tinh: a) loga (a 3 .7a0 b) log^.^a^aX ) c) loga Bai 23. Tinh: a3 .Va a) 4 b ^ 3 2 b) 3log »1 7 c) 7 1 0 ^ 5 d) 102+21og">7 e) gl-log23 ^ 23-41og83 1, 9 21og3 2 + 41og812 ^ 4-log2 3-31og8 5 Bai 24. Tim ca so x, biet: a) logx 2 = 2 b) logx 272 = -6 c)logx243 = 5 d)logx72=-- 5 e)logx272=| f) logx 1 73=-0,l DS: a)x=72 c) x = 3 f)x=l/3 Bai 25. Tim x biet: a) log2 x = - (9 log2 4-3 log2 5) 2 b) log3 x = 1 + 2 log3 5- 4 log3 — 2 V 7 21 c) log1 0 x = -2 + 2 log1 0 —10 - ^° 2 log 1 0 1 0 0 5 d) log3 x + log9 3x + log2 7 x = - e) log2 x + log2 (x + 1) = 3 Bai 26. Cac logarit sau duong hay am? a) log2 5 b) log5 2 c)log02 0,8 d)logl77 5 DS: a) duong b) duong c) duong d) am Bai 27. So sanh cac so sau day: a) log34 va log A b) log0172 va log02 0,34 3 •BDHSG DSGT12/1- 225 Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
c) log 3 - va log 5 - d) 2 ,oge 3 va 3 2 1 5 - B | 4 Bai 28. So sanh: 25 a) log4 5 va log 1 — b) log5 7 va log8 7 16 c) log8 27 va log9 25 d) log1 3 5 6 7 5 va log4 5 75 Bai 29. Chung minh: 5 a) log 6 7>log 7 6 b)2<log 2 3 + log 3 2 < - 2 Bai 30. Chung minh: a) l g & + lg b < lg^^ . Va,b>0 b)7log2 a + 7log2 b < 2^1og2 Va,b> l Bai 31. Chung minh neu logx a,logy b,logz c tao thanh mot cap so cong theo thutudo thi: logb y = 2 ° g a X ° g c Z (0 < x,y,z,a,b,c * 1). loga x + logc z Bai 32. Tinh gon: a) A = log3 6. log8 9. log6 2 b) B = loga b2 + loga 2 b4 c) C = Jlogo.s 4 d) D = 36 ,OB65 +I0 1 -l g 2 e) E = log3 2.log4 3. log5 4. log6 5.logv 6.log8 7 DS: a) 2/3 c) 2 d) 30 e) 1/3 Bai 33.Tinh: a) log4g 32 theo a = log2 14 b)log2 5 15 theo h = log1 5 3 c) lg56 theo c = lg2, d = log2 7 d)log3 0 1 3 5 0 theo x = log3 0 3, y = log3 0 5 Bai 34. Tim f(0) de f (x) = 2X ( 1 ~ cos4x ) lig n tu c tr g n R x Bai 35. Chung minh phuong trinh 3X + 4 X = 9X cd nghiem. H D: Dung ham sd lien tuc hoac dao ham. T' U \ i • \/x+x 2 + x + l , %/5x + l - 1 Bai 36: linh a) hm b) hm x->-l X + 1 x->0 X DS: a) - 2/3 b) 1 n-'-iT T - l , \ r e a x - e b x ... . ln(6x + 1) - ln(2x + 1) Bai 37: Tinh a) hm b) hm — x->0 x x -*0 X DS: a)a- b b) 4 226 -BDHSG DSGTUI1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Bai 38: Tinh v ,. V l + 3x2 -cos2x , N 2%/l + x - 78 - x a) hm - b) hm x-vO x x->0 X DS: a) 7/2 b) 13/12 Bai 39: Tinh \ -i• 3X - 1 , x ,. f x + 4 a) hm b) hm *° Vx + 1 - 1 x^+co^X + 1 DS: a)21n3 b) e 3 Bai 40: Tinh I i a) lim ( l + s in2x)x b) lim ( l + 8x)x *->0+ ' x^0+ V ' DS: a)e 2 Bai 41: Tinh dao ham cap n cua ham so: a)y = 3 x + 1 b)y = ln(x 2 - 4 ) DS:b) (-ir\n - l V . + (-ir\n-lV. ( x -2) n (x + 2) n Bai 42: Tinh dao ham cap n cua ham so: a) y = x. ex b) y = x3 .lnx DS: a) y , n > = ( x+ n) ex Bai 43: Tinh dao ham cap n cua ham so: y = ex sin x n DS: (e x sin x)( n ) = 21 ex sin(x + n- ) 4 Bai 44: Tim khoang don dieu va cue tri cua ham so a) y = x.e2x b) y = x2 lnx c) y - e Bai 45: Tim GTLN, GTNN cua ham sd: 2. 1 a ) y = lnx-V x b)y = lg x + lg x + 2 Bai 46: Tim m dd bat phuong trinh 4.2cos2 x + 2 W v > 2s a) cd nghidm b) cd nghidm vdi moi x Bai 47: Cho tam giac ABC, tim GTNN cua Q = (tan^) 2 7 5 + (tan|) 2 ^ + (tan^) 2Vz ~ > 3X + 4 X + 5X '20Y U U J i H D : Dung bat dang thucGdsi -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
3 Bai 49: Cho a, b, c la cac so ducmg thoa man a + b + c = — 4 Chung minh: 7a + 3b + 7b + 3c + 7c + 3a < 3 . H D: Dung bat dang thuc Cdsi Bai 50: Cho x, y, z la ba so thoa man x + y + z = 0. Chung minh: 73 + 4" + 73 + 4x + 73 + 4x > 6 . H D: Dung bat dang thuc Cdsi Bai 51: Cho 3 so a, b, c cd tong a + b + c =0 Chung minh: 8a + 8b + 8C > 2a + 2b + 2C H D: Dat x = 2a , y = 2b , z = 2C => x, y, z > 0; x + y + z > 3 Bai 52: Chung minh: log2 (x2 + 1) - log2 x > 3x2 - 2x3 , Vx > 0 H D: Dung dao ham Bai 53: Cho 2 sd a, b ma a + b > 0 Chung minh: ' a + b Y a n + b n w < , Vn e N 2 j 2 H D: Dung dao ham 54: Chung minh 4 ^ 'x3 - H D: Dua ve ham phin thuc rieng bidt, ham logarit mdt bdn. Bai 55: Cho x > y > 0. Chung minh: (1 + -) x > (1 + - ) y x y H D: Lay logarit Nepe 2 ve. Bai 54: Chung minh ^J n X < X + ^ . Vx > 0,x * 1 X - l x + x Bai 56: Chung minh bat dang thuc Bai 57: Chung minh bat dang thuc Bai 58: Chung minh bat dang thuc Bai 59: Cho 0 < x, y, z < 1 e 2 x > 2x2 + 2x, Vx > 0 3x >x+l, Vx>0 Chung minh: (2X +2y + 2Z ) f 1 1 1 ^ 81 — + — + — < 2X 2y 2Z Bai 60: Chung minh bat dang thuc vdi n nguydn duong xn .7l-x <^=L=. Vx e(0,l). V2ne 228 -BDHSG DSGT12/1- Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
Cdng t i Sach Thie't b i Giao due ANPH A 50 Nguyen Van Sang P.TSIM, Q.Tan Phu DT: 08.62676463 SACH CO BAN TAI Tp. Ha Noi: Cong tiTNHH Trinh Dau 98 Le Thanh Nghj DT: 04.38680092 Cong ti Quang Loi 32 Gia Ngu. DT: 04.38246605 Cong ti Viet Kim Long 59 Nguyin Khoai, Q.HBT DT: 04.39844721 Nha sach Binh Thuy 67 Nguyin Khoai, Q.HBT DT: 04.39845439 Nha sach Ngoc Hoa 50 Ly Thudng Kiet, Q.HK DT: 04.38258410 Nha sach Diftfng Nguyet 42E Ly Thuong Kiet, Q.HK DT: 04.38264906 Tp. Da Nang Nha sach Phifofng 4 Ly Thai To. DT: 3823421 Nha sach Lam Chau 129 Phan Chu Trinh DT: 0511.3821317 Tp. Ho Chi Minh: Trung tam Sach Giao due Anpha. 225C Ng. Tri Phuong, P.9, Q.5. DT: 08.38547464 Cong ti Thai Khue 48 Nguyin Thj Minh Khai, Q.3 DT: 08.38244422 CH Sach Thie't bj giao due 10A-10B Dinh Tien Hoang, Q.1.DT: 08.38228300 Nha sach 142 Tran Huy Lieu DT: 08.38458295 Va he thdng cac sieu thj sach cua cong ti Phifofng Nam, Fahasa, Van hoa Du Ijch Gia Lai... tren toan quoc. Download Ebook Tai: https://downloadsachmienphi.com Tron Bo SGK: https://bookgiaokhoa.com
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