200BI TP N HC K I(Ly tha v logarit)M rng khi nim lu tha

1.Rt gn cc biu thc sau: a)b) c) ( ) 10.27 3 + (0,2) 4.25 2d) c) (a 4 b 4):(a 2 b 2) d) (x3 + y 6):(x + )e)f)(x.a1 a.x 1). 2.Tnh cc biu thc sau:a) 2 : 2 2 . 25 3 b) 3 38 . 2 . 4 c) 1611a : a a a a d) 213 3a : a . a . ae)5 4 3 2x . x . xf) 5 3ba.abg) 5 1 5 25 33 . 26+ ++h) 121212 3 ) 2 3 ( ) 2 3 ( 2 3

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+ +

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+ k) () 0,75 + ( ) 4/3l) 2 4 2 1 2 32 . 2 . 4 +m) 2 2 1 2 2 2 15 ). 5 25 ( +3.Cho hai s a ,b > 0.Tnh cc biu thc sau:a) 24343) a 3 a 2 ( + b) ) a a )( a a )( a a (515254525251 + + c)) 1 a a )( 1 a a )( 1 a a (4 4+ + + + d)a 1) a 1 )( a 1 (a a212121+ + +e) ) a a ( a) a a ( a414341323134++ f) 6 63131b aa b b a++g) ) ab b a )( b a (3 32323 3 + + h)

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+ + +3 33131abba2 : ) b a (

i) 13112 22 24 3 3 4) b a ( :) b a ( a) b a ( b 3) b a (b ab 2 aa ab b a a+1]1

+ ++ ++ + + j)ab 2 ) b a (a ) )ba( 1 (221212 2+ k) .( 1 + ).(a + b + c) 24.Cho bit 4x + 4 x = 23 ,hy tnh 2x + 2 x5.Rt gn cc biu thc sau: a) (a + b ):()b)231 122 2) ab ( :) b a () b a ( 2) b a (b a

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+++++ c)231 1 2a 1 a.a2 2) a 1 (2 a

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+ d)(a4 b) 1 + ( ) 1 e) 1222231 2a 1 a:a2) a 1 (2

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+

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+f) . g) [(a 1 + b 1 )(a + b + 2c)]:[a 2 + b 2 + ]h),_

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+++b11b 1) 1 b (b a a1b a a122i) 22121b a :abab2 1

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+ j) 2121232145414941b bb ba aa a+5.Rt gn cc biu thc sau:a)A = ) 5 2 )( 25 10 4 (3131313131+ + b) B = 21212121y xx . y y . xc) C = abb a) b a )( b a (212143434343+ d) D = 221212121212323a xa x. ) ax (a xa x111]1

111]1

+e) E = ) b a ( :b ab ab . a ab a414141412121412143111]1

++f) F = 22121121211a aa 3 4 aa 3 a 2a 9 a 411]1

+ +g) G = 11]1

+111]1

++121212121212121212323) b a ( b a :b abb aab ab ah) H = 111]1

+111]1

+21212121232312121b ab ab a ab a.a 3a b a 2 i) I = 352 4 4 2 4 43a a .ab a) b a ( ) b a (a1]1

+ + +j)J =32 3323222 2 33232326 4 2 2 4 62b 2 ) a b ( ab a 2 ) a b () b b a 3 b a 3 a (a1111]1

+ + + + + + k) K = 2(a + b) 1.( )122121 a bab . 14 b a 1 _ 1+ 1 , ]vi a.b > 06.Cho 2 sa = 5 2 10 4 + + v b = 5 2 10 4 + Tnha + b6. Rt gn biu thcA = vix = a bb a _+ ,a < 0 ;b < 07.Cho 1x 2. Chng minh rng: 2 1 x 2 x 1 x 2 x + +8.Rt gn cc biu thc sau: a)212122321212a aa 1a2a aa a+ b) : c)21212121b ab a:ab 2 b ab a +

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+ +d)) a b .(b ab ab ab a21212121212121212121 111]1

++e)

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++

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1 a1 a1 a1 a.a 212a2f )1212323) b a () ab (1b ab ab ab 2

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++++g) 121212323b ab a. abb ab ab ab a

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+

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++

h)313132323231313232313132b ab ab b a ab ab b a ab a+ ++ +9**.Rt gn cc biu thc sau:a) 2121121211a aa 2 3 aa 2 aa 4 a++ +++b)32343432232323434a aa 2 a 2 3 a 3a 2 a 5a 4 a 25 + c) 2121121211a 2 aa 2 5 a 2a aa a+ ++ d)2121121211a 3 aa 9 aa 5 aa 10 3 a+ +e)2121121211a 3 aa 15 2 aa 5 aa 25 a +++ f)2121121211a 3 aa 12 1 aa 4 a 3a 16 a 9+ +10.Cho ba s dng tho a + b = c . Chng minh rng :323232c b a > + 11.Cho a,b,c l di cc cnh ca mt tam gic ,chng minh rng nu c l cnh ln nht th: 434343c b a > +12.Cho a ,b 0 v m ,n l hai s nguyn dng tho m n . Chng minh rng :n1n nm1m m) b a ( ) b a ( + +13.Cho f(x) = a)Chng minh rng nu a + b = 1thf(a) + f(b) = 1b) Tnh tngS = f() + f() + + f() + f()14.Tm min xc nh ca cc hm s sau:a) y = (x2 4x + 3) 2 b) y = (x3 3x2 + 2x)1/4c) y = (x2 + x 6) 1/3 d) y = (x3 8)/315.So snh cc cp s sau:a) 2 / 52

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v3 / 102

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b) 22

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v35

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c) 4 / 1053

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v2 / 574

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d) 376

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v287

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e) 56

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v25

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f) 252

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v353

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LOGARIT1.Tnha) 3216 4 log b) 3313 27 logc) 5232 8 log d)3aa a log e) log3(log28)2.Tnh a) 3 log82b)2 log749 c)10 log 3525d)7 log 2264 e) 3 log 224+f)8 log 31010g)(5 log 32) 25 , 0 ( h) 7 log15 log16 849 25 + h) 4 log21391

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3. Chng minh rng 51315 log3

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2b logb aa4.Rt gn cc biu thc sau:a)36 log . 3 log36 b) 81 log . 8 log43 c) 325 22 log .51logd) e) lgtg1o + lgtg2o+ + lgtg89of) 331313145 log 3 400 log216 log 2 + 5.Cho log23 = a ; log25 = b .Tnh cc s sau : log2 ,log23135 , log2180 ,log337,5 ,log3, log1524 , 30 log106.a)Cho log53 = a,tnh log2515b) Cho log96 = a, tnh log18327.Cho lg2 = a , log27 = b,tnh lg568.Cho log615 = a ,log1218 = b , tnhlog25249.Cho log257 = a ,log25 = b hy tnh 849log3510. Chng minh rng log186 + log26 = 2log186.log2611.a)Cho lg5 = a ,lg3 = b tnhlog308b) Cho log615 = a ,log1218 = btnh biu thc A = log2524c) Cho log45147 = a ,log2175 = b, tnh biu thc A = log497512. Cho log275 = a , log87 = b , log23 = c .Tnh log635theo a,b,c13.Cho log23 = a , log35 = b , log72 = c .Tnh log14063theo a,b,c14.Cho a2 + b2 = 7ab a > 0, b > 0,chng minh rng :lg() = ( lga + lgb )15.Cho a2 + 4b2 = 12ab a > 0, b > 0,chng minh rng: lg(a + 2b) 2lg2 = ( lga + lgb )16.a)Cho x2 + 4y2 = 12xy x > 0,y > 0,chng minh rnglg(x + 2y) 2lg2 = (lgx + lgy)b)Cho a,b > 0 tho mn 4a2 + 9b2 = 4ab v s c > 0,1,chng minh rng :logc =17.Cho log1218 = a , log2454 = b ,chng minh rngab + 5(a b) = 118.Cho logaba = 2, tnh biu thc A = logab18. Chng minh rng :a) a log b logc cb a b) = 1 + logabc) logad.logbd + logbd.logcd + logcd.logad =19.Cho a,b,c,N > 0,1 tho mn:b2 = ac . Chng minh rng : 19.Cho x lg 1110 y,y lg 1110 z . Chng minh rng :z lg 1110 x

20.So snh cc cp s sau:a) log43vlog56 b) 5 log21v3 log51c)log54vlog45d) log231vlog527e)log59vlog311f)log710vlog512 g)log56vlog67 h)logn(n + 1)vlog(n + 1)(n + 2)20.Tm min xc nhca cc hm s sau:a)y = log6 b) y = c) y = 21.a) Cho a > 1. Chng minh rng :loga(a + 1) > loga +1(a + 2)b)T suy ralog1719 > log1920Phng trnh m1.Gii cc phng trnh sau:a) 22x 4 = 5 x 3 x24 +b)3x 2 = 2 c)0,125.42x 3 = 2)82(d)2 x2 x 41 x1 x81 .9127++e) 2x.5x 1= .102 x f) 2x.3x 1.5 x 2 = 12 g) 3 x) 1 x (+= 1h) 1 x 22) 1 x x (+ = 1 i) ()x 2 = 1j) 2x 4 2) 2 x 2 x (+ = 12.Gii cc phng trnh sau:a)500 8 . 5x1 xxb)36 8 . 31 xxx+c) 9x 2x + 1 =2x + 2 32x 1d)2 xx8+ = 36.32 x 3.Gii cc phng trnh sau:a) 2x 4x 1 = 1b) 5x 1 + 5 x+3 = 26c)92x 32x 6 = 0 c)4x + 1 16x = 2log48 d)2x 1 22 x = e)3x + 1 + 32 x = 28f)= 5 g)8x + 18x = 2.27x h)0 12 2 8x 3 x 3x2 + +i)4 3 2 3 2x x + +j)(7 + 4)x + 3(2 )x + 2 = 0 k)14 ) 48 7 ( ) 48 7 (x x + + l)6 2 . 5 42 x 1 x 2 x x2 2 + + m) 32x + 1 = 3x + 2 +n)6 2 . 4 2x cos x sin2 2 +o) (26 + 15)x + 2(7 + 4)x 2(2 )x = 14.Gii cc phng trnh sau: a) 3.4x +2.9x = 5.6xb)6.9x 13.6x + 6.4x = 0 c)4.9x 6x = 18.4x d) 5.36x = 3.16x + 2.81x e) 3.2 2lnx+ 4.6lnx 4.3 2lnx = 0

f)3x + 1 + x 2x + 1 = 0 g) x x 1 x x2 . 3 4 4+ + h)1 22 10 25+ +x x x i)2 2 22 1 2 1 215 . 34 9 25x x x x x x + + + j) 5.32x 1 7.3x 1 += 0k) (3 + )x + 16(3 )x = 2x + 3 5.Gii cc phng trnh sau:a)3x = 13 2xb) 3x = x + 11c)4x 3x = 1d)2x = 3x/2 + 1 e)2x = 3x 5 f)3x = 5x/2 + 4g) 3x1 =34 5x1h)52x = 32x + 2.5x + 2.3x i) 1 + 26x + 24x = 34x

h) (2 )x + (2 + )x = 4x 6.Gii cc phng trnh sau: a) 3.4x + (3x 10).2x+ 3 x = 0 b) 9x + 2(x 2).3x+ 2x 5 = 0 c) 25x 2(3 x).5x + 2x 7 = 0 d) x2 (3 2x )x + 2 2x +1 = 0 e) 3.25x 2 + (3x 10).5x 2+ 3 x= 0f) 2x1 x x22= (x 1)2 f) (4x 1)2 + 2x + 1(4x 1) = 8.4x 7. a)Chng minh rng : = 2 b)T gii phng trnh :(cos720)x (cos360)x = 2 x8.Tm m phng trnh: m.2x + 2 x 5 = 0 c 1 nghim duy nht9.Tm m phng trnh 4x m.2x+1 + 2m = 0 c 2 nghim x1,x2

tho x1 + x2 = 310.Tm m cc phng trnh sau c nghim : a) m.2x + (m + 2)2 x + m + 2 = 0 b) m.3x + m.3 x = 8 c) (m 1)4x + 2(m 3)2x + m + 3 = 0d) (m 4).9x 2(m 2).3x + m 1 = 0 e) 0 3 3 ). 1 m ( 9 ) 1 m (2 2x x + + + +f)0 m 3 . m 3x cos x sin2 2 + +11.Tm m phng trnh : (m + 3)4x + (2m 1)2x + m + 1 = 0 c 2 nghim tri du12.Tm tt c cc gi tr ca m sao cho bt phng trnh sau c nghim ng x 0 : m.2x+1 + (2m + 1)(3 )x + (3 + )x ++ d) 2 x31+

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> 3 x e) 2 x6 x 5 x31312 + +> e) x 5 2x 5 652+

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< f) 3 x 2 2 x 2 x 44 42

g) 4x 3.2x + 2 3i) 4x2 + x 1 x3 x . 3++ < 2.2 xx . 3 + 2x + 6j) 4x2 + x12 x 8 2 x 2 . 3 22 2 2x 2 x 1 x+ + > ++k)4 x 4 x x x 29 . 9 3 . 8 3+ + + > 0l) 12 22 ) 1 5 (+ + + + +x x x x < x x + 2) 1 5 ( 3m)1n) + 21+ x > 5 o)1 x1 x2) 1 x 2 x (++ 1p) ( )x 1 ( )x > 2log482.Cho bt phng trnh : 4x 1 m(2x +1) > 0a)Gii bt phng trnh khi m = 16/9b)Xc nh m bt phng trnh tho mn x R3*.Tm m :a)m.4x + (m 1)2x + 2 + m 1 > 0 xb)m.9x (2m + 1)6x 4x < 0 x [0;1]c)4x - m2x + m + 3 < 0 c nghimd) (m 1).4x + 2(m - 3)2x + m + 3 < 0 c nghim4*.Cho 2 bt phng trnh :x1x23131

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+

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> 12(1)v2x2 + (m + 2)x + 2 3m ++ < + + +2 ) 2 x ( log) 12 lg(7.2 ) 1 2 lg( 2 lg ) 1 x (xx 1 x(1) b)Tm cc gi tr ca m phng trnh m.22x (2m + 1)2- x m + 4 = 0 c hai nghim phn bit x1,x2 (x1 < x2 )saocho x1 nm ngoi v x2 nm trong khong nghim ca h (1)

13.a)Gii bt phng trnh> 3 (1) a l tham s > 0; 1 b)Tm cc gi tr ca m sao cho mi nghim ca (1) cng l nghim ca bt phng trnh : 1 + log5(x2 + 1) log5(x2 + 4x + m) > 0(2)14.Vi gi tr no ca a th bt phng trnh log2a +1(2x - 1) + loga(x + 3) > 0 c tho mn ng thi ti x = 1 v x = 415.Gii bt phng trnh: (2 + )( 1) (+ 2)logx16.Cho h phng trnh ' + 0 ay y x0 y log x log212 3323 a l tham s a)Gii h khi a = 2 b)Xc nh a h c nghim.Gii cc h phng trnh : a)'+ + ++ 6 y 3 x 3 y x) xy ( 2 3 92 23 log ) xy ( log2 2 b)' + +4 y log x log 25 ) y x ( log2 42 22