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<span class='text_page_counter'>(1)</span>ÔN TẬP PHƯƠNG TRÌNH, BẤT PHƯƠNG TRÌNH MŨ VÀ LOGARÍT. Haûi 1.

<span class='text_page_counter'>(2)</span> 2.

<span class='text_page_counter'>(3)</span> 1.y= y=e. -. 1 3 2x 3. e 2x. 2x =>y'=(- )'e 3. -. 2x 3. =-. 2 3 3 e 2x. sin 2 x. 2)y=e 2 sin 2 x 2 2 y'=(sin x)'e =2sinxcosxsin x=sin xsin2x 2. y= (3x-2) 2(3x-2)' 6 y'= 3 = 3 3 3x-2 3 3x-2 3. 3.

<span class='text_page_counter'>(4)</span> Bài 2: Tìm cực tri 1)y=x2ex D=R y’=2xex+x2ex=ex(x2+2x); y’=0x2+2x=0x=0 hoặc x=-2 x - -2 0  y’ + 0 - 0 + y VẤN ĐỀ II: PHƯƠNG TRÌNH,BPT MŨ VA LÔGA RÍT 4.

<span class='text_page_counter'>(5)</span> VẤN ĐỀ II: PHƯƠNG TRÌNH,BPT MŨ VA LÔGA RÍT x 2 -6x-9. Bai 2: Giai 1)(2/3) 2 -2 x 2 -6x-9  (2/3) <( ) 3. <9/4. x2-6x-9>-2( do 0<2/3<1) x2-6x-7>0x<-1 hoặc x>7 2)log2x +log4x + log8x = 11 Đk:x>0 Ptlog2x +(1/2)log2x + (1/3)log2x = 11 (11/6)log2x=11log2x=6x=26x=64 5.

<span class='text_page_counter'>(6)</span> 3)log 2  x- 2  – 2 < 6log1/8 3x-5 ĐK: x>2; khi đó bpt  1 2.  log 2  x- 2  -log 2 2 2 < 6log 2-3 (3x-5) x- 2  log 2 <- log 2 (3x-5) 4 x- 2 x- 2  log 2 (3x-5)<0  (3x-5)<1 4 4 3x2-11x+10<43x2-11x+6<02/3<x<3 So đk ta có: 2<x<3 x. x 2. 2  4 =3 +1x 4)2x =3x/2 +1 x x x 3 2 1 2 3 2 1 2  1=( ) +( )  ( ) +( ) -1=0 4 4 4 4. 6.

<span class='text_page_counter'>(7)</span> x/2 x/2  4 =3 +1 4)2 =3 +1 3 x2 1 x2 3 x2 1 x2  1=( ) +( )  ( ) +( ) -1=0 4 4 4 4 3 x2 1 x2 1 3 x2 3 1 x2 1 hs y=( ) +( ) -1;D=R=>y'= [( ) ln +( ) ln ]<0;x  R 4 4 2 4 4 4 4 x. x/2. =>f(x) giảm trên R; mà f(2)= 1-1=0=>x=2 là ngh duy nhất. 5)9x-1 +3=36.3x-3 (1/9).9x+3=(36/27).3x9x-12.3x+27=03x=9 hoặc 3x=3 x=2 hoặc x=1 6)log2(x-1)-(1/2)log2(x2- 4)=1 Đk:x>2; khi đó pt: 7.

<span class='text_page_counter'>(8)</span> 6)log2(x-1)-(1/2)log2(x2- 4)=1 Đk:x>2; khi đó pt: 2log2(x-1)-log2(x2-4)=2 log2(x-1)2-log2(x2-4)=2 2 (x-1)2 (x-1)  log 2 2 =log 2 2 2  2 =4 x -4 x -4 3x2+2x-17=0  x= -1±2 13 ; so dk => ngh:x= -1+2 13 3 3 7)2.9x-5.6x+3.4x>0 9 x 3 x  2.( ) -5.( ) +3>0  4 2 1 x2 -2x-11 8)( ) <8 2.  3 x  ( 2 ) <1  x<0    x>1  ( 3 )x > 3  2 2 8.

<span class='text_page_counter'>(9)</span> 1 x2 -2x-11 8)( ) <8 2 2  2-x +2x+11 <23 -x2+2x+11<3-x2+2x+8<0x<-2 hoặc x>4 9)49x -7 x+1 +6=0  x=0 x x x x 49 -7.7 +6=07 =1 hoặc 7 =6    x=log 7 6. 11)lgx +1>2lg(x+1). Đk: x>0. Bptlg10x>lg(x+1)2 10x>(x+1)2x2-8x+1<0 12)4x+10x=2.25x 25 x 5 x 5 x 5 x 1  2.( ) -( ) -1=0  ( ) =1 hoac ( ) =- (loai)  x=0 4 2 2 2 2 9.

<span class='text_page_counter'>(10)</span> 13)lg(x-2)+lg(x-3)=1-lg5; đk: x>3. Khi đó pt: lg(x-2)(x-3).5=lg105(x-2)(x-3)=10 5x2-25x+20=0x=1(loại); x=4(nhận) 14)log(x-3)=2-log50; đk: x>3 ; khi đó: pt(x-3)50=log100 50(x-3)=100x=5(nhận) 15)log2(x2-3)-log2(6x-10)+1=0 Đk:x>3; khi đó Pt:log2(x2-3)4=log2(6x-10) 4(x2-3)=6x-104x2-3x-10=0 x=2(nh);x=-5/4(loại) 10.

<span class='text_page_counter'>(11)</span>  5 x 5 ( 2 ) = 2 5 x 25 x  5-7( ) -2( ) =0    2 4  ( 5 )x 1  2 3x 2 18)log1/2 ( -1)=log1/2 (2x -5x+4) 2 16) 5.4x -7.10x+ 2.25x=0.  x=1  x 0 . Đk: 3x/2 -1>0x>2/3; khi đó: Pt3x/2 -1=2x2-5x+4 4x2-13x+10=0x=2(nh) hoặc x=5/4(nh). 19) 9. x 2 +1. x2. -3(3 +4)<0 11.

<span class='text_page_counter'>(12)</span> 19) 9. x 2 +1. x2. -3(3 +4)<0. x2. x2.  9. 9 -3.3 -12<0 4 4 4 4 x2 2  0<3 <  x <log 3  - log 3 <x< log 3 3 3 3 3 20)log1/2(x2 -4x+6) < - 2 x2-4x+6>(1/2)-2x2-4x+2>0.  x<2- 2   x>2+ 2 21)4x+3/2+9x =6x+1 12.

<span class='text_page_counter'>(13)</span> 21)4x+3/2+9x =6x+1 8.4x+9x -6.6x=0 9 x 3 x 3 x   8+( ) -6.( ) =0 ( ) 4  4 2 2  22)(1/2)2x -x-10 >16  ( 3 ) x 2 1 2x -x-10 1 -4   2 ( ) >( ) 2 2 2x2-x-10<-42x2-x-6<0-3/2<x<2 2. 2. 23)log2(x -3)- log1/2(x-1) < 3. 13.

<span class='text_page_counter'>(14)</span> 23)log2(x -3)- log1/2(x-1) < 3 log2(x -3)+ log2(x-1) <log223;ñk: x>3 log2(x -3)(x-1)< log1/28 (x-3)(x-1)<8x2-4x-5<0-1<x<5 So dk ta co: 3<x<5 |x 2 -x|. 24)1<5. 25. 2.  x -x  0 2 |x -x|>0  2  2  x -x 2 |x -x| 2 x 2 -x  2 . 14.

<span class='text_page_counter'>(15)</span> |x 2 -x|. 24)1<5. 25. 2.  x -x  0 2 |x -x|>0  2  2  x -x 2  |x -x| 2 x 2 -x  2  x 0 va x 1  -1  x 2. x 0 va x 1  2 x -x-2 0 x 2 -x+2 0(dung ) . 25)lg2x – lgx2 = lg23 – 1 lg2x –2lgx = lg23 – 1;x>0 15.

<span class='text_page_counter'>(16)</span> 26)52x+1 > 5x +4  x 4 5 <- (loai)  2x x  x>0 5.5 -5 -4>0 5  x  5 >1. 27)2log8(x- 2) + log1/8( 2x – 7) > 2/3 đk:x>7/2 Bptlog8(x- 2)2 – log8( 2x–7)>2/3 2 2 (x-2)2 2 (x-2)  log 8 >  >8 3 2x-7 3 2x-7. x2-12x+32>0x<4 hoặc x>8 So với đk ta có: 7/2<x<4 hoặc x>8 16.

<span class='text_page_counter'>(17)</span> 28)log1/32x – 2log1/3x -3 >0 Đk:x>0  log1/3 x<-1 bpt     log1/3 x>3.  x>3  0<x<1/27 . 29)32x+3 – 4.3x+1 + 1 = 0  x 1 3 = 3  27.32x-12.3x+1=0    3x = 1  9 30)log2[log1/2(x2 -2)]<1 Đk: log1/2(x2-2)>0; khi đó bpt:.  x=-1  x=-2 . 17.

<span class='text_page_counter'>(18)</span> 30)log2[log1/2(x2 -2)]<1 Đk: log1/2(x2-2)>0; khi đó bpt: log 1 (x 2 -2)>0  2  2 log (x  1 -2)<2  2 x 2 -2<1 x 2 -3<0  2   x -2>0   2 9  x 2 -2>1/4 x - >0  4 . 3   - 3<x<- 2   3 <x< 3  2. 18.

<span class='text_page_counter'>(19)</span> 31)log2(x2 +7x+3)<-log1/2(2x-1)  log2(x2+7x+3)<log2(2x-1). x 2 +7x+3>0  2  x +7x+3<2x-1. x 2 +7x+3>0  2 VN x +5x+4<0 . 32)x(lg5 -1)=lg(2x+1)–lg6 xlg(5/10)=lg(2x+1)–lg6 -lg2x=lg(2x+1)–lg6 lg2x(2x+1)=lg62x(2x+1)=6 4x+2x-6 =02x=2 hoặc 2x= -3x=1 33)log2(x+3) = log3(x+8) Đặt t=log2(x+3)x=2t-3 19.

<span class='text_page_counter'>(20)</span> 33)log2(x+3) = log3(x+8) Đặt t=log2(x+3)x=2t-3 Khi đó pt: t=log3(2t+5)3t=2t+5 2 t 1 t  1=( ) +5.( ) 3 3 2 t 1 t Xet hs:y=( ) +5.( ) ;t  R 3 3 2 t 2 1 t 1 y'=( ) ln( )  ( ) ln( )<0; t  R 3 3 3 3 =>hs nghich biến trên R; mà f(2)=1=>t=2 là ngh duy nhất của pt f(t)=1 Ta có t=2x=1 20.

<span class='text_page_counter'>(21)</span> 34)9x +2(x -2).3x + 2x – 5 = 0 Đặt t=3x;t>0; pt: t2+2(x-2)t+2x-5=0; =(x-3)2 t= -1(loại);t=-2x+5 Vậy : 3x=5-2x3x+2x-5=0 Xét hs y=3x+2x-5;D=R y’=3xln3+2>0;x => hs đồng biến trên R Mà f(1)=0 nên x=1 là ngh duy nhất của pt 35)-x.2x = x(3–x)-3.2x. 21.

<span class='text_page_counter'>(22)</span> 35)-x.2x = x(3–x)-3.2x 2x(-x+3) =x(3-x)(3-x)(2x-x)=0.  x=3  x  2 -x=0 Xét hs y=2x-x; D=R +)Nếu x<0=>y>0 mọi x=>pt VN +)Nếu x1=> y’=2xln2-1 2ln2-1=ln4-1>0=>hs tăng=>f(x)f(1)=1>0. +)x[0;1)=>2x 1;-x>-1=>y>0 Vậy pt 2x-x=0 VN=> pt đã cho có ngh duy nhất x=3 36)ln(x+1) +ln(x+3)= ln(x+7) 22.

<span class='text_page_counter'>(23)</span> 36)ln(x+1) +ln(x+3)= ln(x+7). 1 2 37) + =1 4-log 2 x 2+log 2 x 1 2 + =1;t 4;t -2 Đặt t=log2x; pt: 4-t 2+t t2-3t+2=0t=1(nh) hoặc t=2(nh) log2x=1 hoặc log2x=2x=2 hoặc x=4 38)log2x + log3x + log4x = log20x 23.

<span class='text_page_counter'>(24)</span> 38)log2x + log3x + log4x = log20x Đk: x>0 Xét hs y=log2x + log3x + log4x- log20x; x>0 1 1 1 1 1 1 1 1  y'= + + =  + xln2 xln3 xln4 xln20 x  ln3 ln4 ln20  1 1 1 0<ln3<ln4<ln20=> + >0 ln3 ln4 ln20 =>y’>0;x>0=> hs đồng biến Mà f(1)=0=>x=1 là ngh duy nhất của pt đã cho 39)log4log2x+log2log4x=2 24.

<span class='text_page_counter'>(25)</span> 39)log4log2x+log2log4x=2 Đk: log2x>0;log4x>0x>1 Khi đó pt: log4(log2x)+log4(log4x)2=2 log4[(log2x).(log4x)2]=2 (log2x).(log4x)2=16 2. 2.  log 4 x (log 4 x) =16  2log 4 x(log 4 x)2 =16 3.  (log 4 x) =8  log 4 x=2  x=16(nh) 40)log 2 (1+ x )=log 3 x 25.

<span class='text_page_counter'>(26)</span> 40)log 2 (1+ x )=log 3 x. Đk: x>0 Đặt t=log3xx=3t; khi đó pt:. log 2 (1+ 3t )=t  1+ 3t =2t t t t t 1 2 3 2 2 2  1+3 =4  ( ) +( ) =1 t 4 t 4 1 2 3 2 Xet hs:y=( ) +( ) ;D=R 4 4 t t 1 1 2 1 1 1 2 1 y'= ( ) ln( )+ ( ) ln( )<0;x 2 4 4 2 3 3 =>hs nghich biến trên R mà f(2)=1=>t=2 là ngh duy nhất của pt f(t)=1. =>x=9 là ngh pt đã cho 26.

<span class='text_page_counter'>(27)</span> 41)lg(x2-x-6)+x=lg(x+2)+8. Đk: x>3. 2. x -x-6 pt:lg +x-8=0 lg(x-3)+x-4=0 x+2. Xét hs: y=lg(x+3)+x-8=0;x>3 y’=1/xln10 +1>0;x>3=>hs tăng trên TXĐ Mà f(7)=0=>x=7 là ngh duy nhất của pt. 42)32. x+9 x-5. =0,25.125. x+9 5 x-5.  (2 ) 2. 5(x+9) 2 x-5. =2-2 .5 =5. 3.. 3(x+7) x-3. x+7 x-3. x+7 x-3. 3 x-5.   7x+35 3(x+7) 2   x-5 x-3  2 =5  1   5 x-3 . x+7. =1 27.

<span class='text_page_counter'>(28)</span> 43)log4[(x+2)(x+3)] +log4[(x-2)/(x+3)]= 2 Đk:x<-3 hoặc x>2 x-2 Khidopt:log 4 (x+2)(x+3) =2 x+3 2 2 log4(x -4)=2x -4=16.  x= 2 5 (nh) 44)16x = log1/2x. 28.

<span class='text_page_counter'>(29)</span> 45)log2(2x+1)log2(2x+1+2)=2. 46)xlg9 + 9lgx = 6. 29.

<span class='text_page_counter'>(30)</span> 47)x. 3lg 3 x-(2/3)lgx. 3. =100 10. 30.

<span class='text_page_counter'>(31)</span> 48)1+logx+25 = log5(x+2). 49)22x-1+22x-2+22x-3448. 31.

<span class='text_page_counter'>(32)</span> 50)1/(5lgx) /(1+lgx) < 1. 51)log1/3log2x2 > 0. 32.

<span class='text_page_counter'>(33)</span> 52)log2x ≤6 – x. 53)e2+lnx =x+3. 54)log2(3x+1)log3x = 2log2(3x+1) 33.

<span class='text_page_counter'>(34)</span> 55)2. log 3 x 2. .5. log 3 x. =400. 34.

<span class='text_page_counter'>(35)</span> 56)log0.22x –log0.2x -6 ≤ 0. 35.

<span class='text_page_counter'>(36)</span> x. x+1. 4 -2 +8 x 57) <8 1-x 2. 36.

<span class='text_page_counter'>(37)</span> 58)ln|x -2| + ln|x+4| ≤3ln2. 59)8x+1 +8.(0.5)3x +3.2x+3 =125-24.(0.5)x. 37.

<span class='text_page_counter'>(38)</span> x. x. 60)( 6+ 35 ) +( 6- 35 ) =12. 61)hvhk01) log3(x2+x+1) – log3x = 2x – x2. 38.

<span class='text_page_counter'>(39)</span> ĐỀ KIỂM TRA 2008. 39.

<span class='text_page_counter'>(40)</span> 2x. 1.Tinh dao ham: a)y=e sin(2x+1) y’=(e2x)’sin(2x+1)+e2x.[sin(2x+1)]’= =2e2xsin(2x+1)+e2x.2cos(2x+1) =2e2x[sin(2x+1)+cos(2x+1)]. ĐỀ KIỂM TRA 2008 b)y=ln e x +1 x x 1 (e +1)' e y= ln(e x +1)=>y'= = x 2 2(e +1) 2(e x +1) 2a)log 9 (2x-5)+log 3 x-1=1. đk:x>5/2; khi đó pt trở thành: 40.

<span class='text_page_counter'>(41)</span> 2a)log 9 (2x-5)+log 3 x-1=1 đk:x>5/2; khi đó pt trở thành:. log 9 (2x-5)+log 32 ( x-1)2 =1  log 9 (2x-5).(x-1)=log 9 9 (2x-5)(x-1)=92x2-7x-4=0 x=4 (nhận) hoặc x=-1/2(loại). b)3. 2x+1. x. -28.3 +9=0.  3x =9  2x x   3.3 -28.3 +9=0 1 x 3 =  3.  x=2  x=-1 . c)5. log 8 x+2. +5. log 8 x-1. =3 x 41.

<span class='text_page_counter'>(42)</span> c)5. log 8 x+2. +5. log 8 x-1. =3 x. đk:x>0; khi đó pt trở thành: 1 log8x log8 3 x log 8 x 5 .25+ 5 =8 126 log58x log8 x (Dạng m.ax=n.bx). Chia 2 vế cho…  5 =2 5 5 log 5 5 126 5 log8x 5 2  log 8 x=log 5  x=8 ( ) = 2 126 2 126. d)x.2x log 3 (-2x 2 -7x+4)+2>2x.2x +log 3 (-2x 2 -7x+4)(1) Đk: -2x2-7x+4>0-4<x<1/2 (1)  (x.2x -1)[log 3 (-2x 2 -7x+4)-2]>0(2) 42.

<span class='text_page_counter'>(43)</span> d)x.2x log 3 (-2x 2 -7x+4)+2>2x.2x +log 3 (-2x 2 -7x+4)(1) ;-4<x<1/2 (1)  (x.2x -1)[log 3 (-2x 2 -7x+4)-2]>0(2) +)x(-4;0]=>x.2x0 x.2x -1<0 1  2 0<x< x x 2 =>x.2 < <1=>x.2 -1<0 +)x(0;1/2)=>  2 0<2x < 2  Vậy mọi x(-4;1/2)=>x.2x -1<0 Bptlog3(-2x2-7x+4)<2-2x2-7x+4<92x2+7x+5>0 x<-5/2 hoặc x>-1 So đk ta có ngh: -4<x<-5/2 hoặc -1<x<1/2 Đề 2009 sin3x 1.Tinh dao ham a)y=5. 43.

<span class='text_page_counter'>(44)</span> 2009 1.Tinh dao ham a)y=5sin3x y'=(sin3x)'5sin3x ln5=3cos3x5sin3xln5 2 2 b)y=x ln x -1 2.Giải: a)3x+2+7x=4.7x-1+34.3x-1 4 x 34 x x x  9.3 +7 = .7 + .3 7 3 3 x 7 x 7 x 49  .7 = .3  ( ) =  x=2 7 3 3 9 2 x 2 b)log 2 4x+log 2 =8 8. 44.

<span class='text_page_counter'>(45)</span> 2 x 2 Đk:x>0;khi đó pt: b)log 2 4x+log 2 =8 8 (2+log 2 x)2 +2log 2 x-3=8 Đặt t=log2x;pt:. (t+2)2+2t-11=0t2+6t-7=0t=1 hoặc t=-7 log2x=1 hoặc t=log2x=-7x=2 hoặc x=2-7. c)7 logx 49 =343.x 5 Đk: 1x>0 Khi đó, lấy loga cơ số x hai vế, pt log x 7 log x 49 =log x 343.x 5  log x 49.log x 7=log x 7 3 +log x x 5 2logx27-3logx7-5=0  log x 7=-1 1  x=7   (nhan) 5 2/5  log x 7= x=7   2 45.

<span class='text_page_counter'>(46)</span> d)cm: 3xln(ex-1)=2x+5 có ngh duy nhất Đk: x>0; khi đó pt trở thành : 2 x 1 x ( ) +5.( ) -ln(e x -1)=0 3 3 2 x 1 x Xet hs y=( ) +5.( ) -ln(e x -1);x>0 3 3 x 2 x 2 1 x e y'=( ) ln -5.( ) ln3- x <0; x>0(e x -1>0) 3 3 3 e -1 => Hs y giảm trên (0;+). ma lim+ f(x)=+; ma lim f(x)=- x 0. x + . =>pt có duy nhất 1 nghiệm e)32x+1-4.3x+1=0 46.

<span class='text_page_counter'>(47)</span> e)32x+1-4.3x+1=0 x  3 =1  x=0 2x x 3.3 -4.3 +1=0    x  x=-1  3 =1/3 1 f) log 2 (2x+3)-log 4 (x-3)=1 Đk: x>3 2 Khidopt:log 22 (2x+3)-log 4 (x-3)=1 log4[(2x+3)/(x-3)]=1  2x+3 =4 x-3 2x+3=4x-12x=15/2(nh) Đk: x>0 g)53-log5 x =25x Khi do pt:log 5 5 3-log5x =log 5 25x 3-log5x=2+log5xlog5x=1/2x=5 (nhận). 2010. 47.

<span class='text_page_counter'>(48)</span> 2010 a)32x+5=3x+2+2 x  3 =-2/27(loai) 2x x 243.3 -9.3 -2=0    x=-2 x  3 =1/9 x-7 1 1 6 Đk x>3 b)( ) 8 3x 2 x-7 3 -x+7 1   2 6 2 3x  0 1 6 6 x + || 0 + 0 2  0<x 1 -x +7x-6  0   x  x 6 x 2 -3x+2 x 2 +6x+5 2x 2 +3x+7 c)4 +4 =4 +1 Đặt m=x2-3x+2;n=x2+6x+5=>n+m=2x2+3x+1 Pt4m+4n=4m+n+1(4m-1)(4n-1)=0 48.

<span class='text_page_counter'>(49)</span> x 2 -3x+2. x 2 +6x+5. 2x 2 +3x+7. c)4 +4 =4 +1 Đặt m=x2-3x+2;n=x2+6x+5=>n+m=2x2+3x+1 Pt4m+4n=4m+n+1(4m-1)(4n-1)=0 4m=1 hoặc 4n=1m=0 hoặc n=0.  x 2 -3x+2=0  x=1;x=2  2   x=-2;x=-3  x +6x+5=0 1 x-1 2 3 d)log 27 (x -5x+6) = log 3 +log 3 (x-3)2 2 2 2 x -5x+6>0 x<2 hoac x>3  1<x<2   dk: x-1>0  x>1   x>3 (x-3)2 >0  x 3   Khi đó pt 49.

<span class='text_page_counter'>(50)</span> 1 x-1 d)log 27 (x -5x+6) = log 3 +log 3 (x-3)2 2 2 x 2 -5x+6>0 x<2 hoac x>3  1<x<2   dk: x-1>0  x>1   x>3 (x-3)2 >0  x 3   1 x-1 2 Khi đó pt log 3 (x 2 -5x+6)3 = log 1/2 +log (x-3) 3 3 3 2 2 x-1 2  log 3 (x -5x+6)=log 3 +log 3 (x-3) 2 2 x-1 2  log 3 (x -5x+6)=log 3 (x-3) 2  2(x 2 -5x+6)=(x-1)(x-3)2 2 (x-3)[2(x-2)-(x-1)(x-3)]=0  x=3+ 2  x=3(loai)  2   x=3- 2  x -6x+7=0 2. 3. 50.

<span class='text_page_counter'>(51)</span> log 25 81 x 2 -6x+8 x 2 -6x+8 e)log 2  5 dk: >0 x+1 x+1 1 log 2 34 x 2 -6x+8 2 5 log  5 Khi đó pt 2 x+1 2 2 x -6x+8 x -6x+8 log 5 3  log 2 5  log 2 3 x+1 x+1 2 2 x -6x+8 x -14  8  0 x+1 x+1. <=> -1<x0 hoặc x14. 1 2 f)log 3-4x2 (9-16x )=2+  3-4x >0 2 log 2 (3-4x ) dk: 3-4x 2 1  9-16x 4 >0  4. 3-4x 2 >0  2 3-4x 1  51.

<span class='text_page_counter'>(52)</span> 1 f)log 3-4x2 (9-16x )=2+ 2 log 2 (3-4x ) 3-4x 2 >0 2   3-4x >0 2 dk: 3-4x 1   2 3-4x 1  9-16x 4 >0  4 2 2 Khi do pt:log 3-4x2 (9-16x )=log 3-4x2 (3-4x ) +log 3-4x2 2 4.  log 3-4x2 (9-16x 4 )=log 3-4x2 (3-4x 2 )2 .2 9-16x4=2(3-4x2)2(3-4x2)(3+4x2-6+8x2)=0  3-4x 2 =0(loai) 1   2 1  x=  (nh) x  2  4 52.

<span class='text_page_counter'>(53)</span> 2.x y>0;cm:2011lnx-2010lny ln(2011x-2010y) Xét nN*;n>1 x,y như gt. Theo bđt cosi: n n n n n n n n n(n-1) x +(n-1)y =x + (y +y +...+y ) n x y       n-1so hang. xn+(n-1)ynnxyn-1 xnyn-1[nx-(n-1)y] ln(xn)ln{yn-1[nx-(n-1)y]} nlnx(n-1)lny+ln[nx-(n-1)y] nlnx -(n-1)lny ln[nx-(n-1)y] Với n=2011=>đpcm 53.

<span class='text_page_counter'>(54)</span> ĐỀ 2011 1. 32+x-32-x=80.  5 2.    3. x+1.  9     25 . x 2 +2x+1. 3.log 2 (x+1)+2=log 4. 9. log 1 (x+1) 3. 2. =5. log 1 (2x +1) 5. 2.  5 =   3. 9. 4-x +log 8 (4+x)3 5.. 1. log 1 3. x. 6. 4 -2. x+1. 1 > 2x 2 -3x+1 log 1 (x+1) 3. x2. +4 0 54.

<span class='text_page_counter'>(55)</span> ĐỀ 2011 1. 32+x-32-x=80. 9  9.3 - x -80=0 3 x. 9.9x-80.3x-9=0 x.  3 =9  x 1  3 = 9  5 2.    3. x+1.  9     25 . x=2. x 2 +2x-11.  5 =   3. 9. 55.

<span class='text_page_counter'>(56)</span>  5 2.    3. x+1.  5    3.  9     25 . x 2 +2x-11. x+1-2(x 2 +2x-11).  5 =   3.  5 =   3. 9. 9. x+1-2x2-4x+22=92x2+3x-14=0 x=2 hoặc x= -7/2. 3.log 2 (x+1)+2=log. 2. 4-x +log 8 (4+x). 3. Đk: -1<x<4 56.

<span class='text_page_counter'>(57)</span> 3.log 2 (x+1)+2=log. 2. 4-x +log 8 (4+x). 3. Đk: -1<x<4; khi đó:. pt:log 2 (x+1)+2=log 2 (4-x)+log 23 (4+x). 3.  log 2 (x+1)+2=log 2 (4-x)+log 2 (4+x) log2(x+1).4=log2(16-x2) (x+1)4=16-x2x2+4x-12=0. 4. 9. log 1 (x+1) 3. =5. log 1 (2x 2 +1). 2 -log 3 (x+1). pt: (3 ). x=2(nh);x=-6(loai). 5. =(5). ;đk: x>0 -log 5 (2x 2 +1). 57.

<span class='text_page_counter'>(58)</span> 4. 9. log 1 (x+1) 3. =5. log 1 (2x 2 +1) 5. 2 -log 3 (x+1). pt: (3 ). ;đk: x>-1. =(5). -log 5 (2x 2 +1). -2. log 3 (x+1) log 5 (2x     3  =5. 2. +1).  . -1. 1 1 = 2 (x+1) =(2x +1)  2 (x+1) 2x +1 2 2 2 -2. 2. -1. (x+1) =(2x +1)x -2x=0x=0(nh);x=2(nh) x=0(nh);x=1/2(nh). 5.. 1. log 1 3. 1 > 2x 2 -3x+1 log 1 (x+1) 3 58.

<span class='text_page_counter'>(59)</span> 5.. 1. log 1. 1 > 2x 2 -3x+1 log 1 (x+1) 1. 3. 1  < log 3 2x 2 -3x+1 log 3 (x+1)  a>0 va b>0 va a>b 1 1  Taco : <   a<0 va b<0 va a>b a b  a<0 va b>0 log 3 (x+1)>0 x+1>1   2 TH1: log 3 2x 2 -3x+1>0  2x -3x+1>1  2 log 2x -3x+1>log 3 (x+1)   3 2  2x -3x+1>x+1 3. 59.

<span class='text_page_counter'>(60)</span> x+1>1  x>0  2   2x -3x+1>1  x<0  x>1/2  x>5   2 2 2 2x -3x+1>x+1 2x -3x+1>x +2x+1   log 3 (x+1)<0  TH2: log 3 2x 2 -3x+1<0  2 log 2x -3x+1  log 3 (x+1)  3 -1<x<0 -1<x<0  2   2x -3x+1<1  0<x<1/2 (VN)   2 2 2x 3 x+1  x  1 2 x -3x+1>x+1   60.

<span class='text_page_counter'>(61)</span> log 3 (x+1)>0 TH3:  2 log 2x -3x+1<0  3 x>0 x>0   2 2   2x -3x+1<1  2x -3x+1<1  2x 2 -3x+1>0 2x 2 -3x+1>0  . 1 3 Vay:S=(0; )  (1; )  (5;+) 2 2 x. x+1. x. x2. 6. 4 -2. 1   0<x< 2   1<x< 3  2. x2. +4 0 x+1.  4 +4 2 x x2 x x2 Theo bdt cosi: 4 +4 2 4 4 61.

<span class='text_page_counter'>(62)</span> x. x2. x. Theo bdt cosi: 4 +4 2 4 4 x. x2. Hay: 4 +4 2.2 Ma : 2. x 2 +x+1. x+1. x 2 +x. =2. x2. x 2 +x+1 x2. 2 ; x Nen:4 +4 2x+1 ; x 2 x x 4 =4 Vay bpt    x=0 2 x  =0 x 2 x2 C2:Dat t=2 >0; ta duoc: t -2t+4 0 2 x2  (t-1) +(4 -1) 0 2 x2 x x+1 x2 Vi (t-1) 0;(4 -1) 0;x,t 6. 4 -2 +4 0 t-1=0 Nen bpt   x2  x=0 4 -1=0 62 x.

<span class='text_page_counter'>(63)</span> 1.log 2013 x-log 2012 x<1  log 2013 2012.log 2012 x-log 2012 x<1  log 2012 x(log 2013 2012-1)<1 1  log 2012 x> (log 2013 2012-1<0) log 2013 2012-1 1  log 2012 x>  log 2012 x>log 2012 2013 2012 2013 log 2013 log 2012 2013 2013  x>2012 2013 x-1. 2. 3.2 -2. x 2 -x. -4. x-1. 0 63.

<span class='text_page_counter'>(64)</span> x-1. x 2 -x. x-1. x-1. 2. x-1. x-1. 2. 2. 3.2 -2 -4 -1 0 x-1 x-1 x-1 x 2 -x  (2.2 -4 -1)+(2 -2 ) 0  -(2 -1) +2 (1-2 2. ) 0. x-1. (x-1)2. -1] 0. x-1. (x-1)2. -1] 0.  -(2 -1) -2 [2 x-1. x 2 -2x+1.  (2 -1) +2 [2. Ta thay VT 0; x x-1. 2. (2 -1) =0 Vay bpt    2 x-1 (x-1) -1] 0 2 [2. x-1. 2 =1  (x-1)2  x=1 =1 2 64.

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