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CC PHNG PHP GII PHNG TRèNH- BT PHNG TRèNH- H M- LễGARIT CHNG I: PHNG PHP GII PHNG TRèNH- BT PHNG TRèNH- H M CH I:PHNG TRèNH M BI TON 1: S DNG PHNG PHP BIN I TNG NG I Phng phỏp: Ta s dng phộp bin i tng ng sau: a = a > f ( x) g ( x) a =a < a hoc ( a 1) f ( x ) g ( x ) = f ( x ) = g ( x ) II VD minh ho: ( VD1: Gii phng trỡnh: + x x ) sin ( = + x x2 ) cos x Gii: Phng trỡnh c bin i v dng: < x < 2(*) + x x > x x = 0(1) 2 + x x sin x + cos x = sin x + cos x = 2(2) ( )( ) tho iu kin (*) cos x = sin x x + ữ = x + = + 2k x = + 2k , k Z Gii (2): sin x + 2 3 nghim tho iu kin (*) ta phi cú: < + k < ữ < k < ữ k = 0, k Z ú ta nhn c x3 = 6 Vy phng trỡnh cú nghim phõn bit x1,2 = ; x 3= Gii (1) ta c x1,2 = VD2: Gii phng trỡnh: ( x 3) x x + ( = x2 x + Gii: Phng trỡnh c bin i v dng: ( x 3) ) x2 + x x x + 2 = ( x ) x2 + x = ( x 3) 2( x + x 4) x = x = x = < x x < x = 3x x + = x + x x x + 10 = Vy phng trỡnh cú nghim phõn bit x=4, x=5 BI TON 2: S DNG PHNG PHP LễGARIT HO V A V CNG C S I Phng phỏp: chuyn n s s m lu tha ngi ta cú th logarit theo cựng c s c v ca phng trỡnh, ta cú cỏc dng: Dng 1: Phng trỡnh: < a 1, b > a f ( x) = b f ( x ) = log a b Dng 2: Phng trỡnh : a f ( x ) = b g ( x ) log a a f ( x ) = log a b f ( x ) f ( x ) = g ( x ).log a b f ( x) = log b b g ( x ) f ( x).log b a = g ( x) hoc log b a II VD minh ho: VD1: Gii phng trỡnh: x2 x = 2 Gii: Ly logarit c s hai v phng trỡnh ta c: log 2 x x = log x x = log x x + log = , Ta cú = + log = log > suy phng trỡnh cú nghim x = log VD2: Gii phng trỡnh: x x.8 x = 500 Gii: Vit li phng trỡnh di dng: x x = 500 x x x x = x x =1 Ly logarit c s v, ta c: x x x x x x x log ữ = log + log ữ = ( x ) log + log 2 = x x = ( x 3) log + ữ = x = x log Vy phng trỡnh cú nghim phõn bit: x = 3; x = log Chỳ ý: i vi phng trỡnh cn thit rỳt gn trc logarit hoỏ BI TON 3: S DNG PHNG PHP T N PH- DNG I Phng phỏp: Phng phỏp dựng n ph dng l vic s dng n ph chuyn phng trỡnh ban u thnh phng trỡnh vi n ph Ta lu ý cỏc phộp t n ph thng gp sau: ( k 1) x .1a x + = Dng 1: Phng trỡnh k + k 1a ( ) k k Khi ú t t = a x iu kin t>0, ta c: k t + k 1t 1t + = M rng: Nu t t = a f ( x ) , iu kin hp t>0 Khi ú: a f ( x ) = t , a f ( x ) = t , , a kf ( x ) = t k f ( x) = V a t x x Dng 2: Phng trỡnh 1a + a + = vi a.b=1 x Khi ú t t = a x , iu kin t0, suy b t Dng 3: Phng trỡnh 1a x + ( ab ) + 3b x = ú chia v ca phng trỡnh cho b x >0 x 2x x a a ( hoc a , ( a.b ) ), ta c: ữ + ữ + = b b x 2x x a t t = ữ , iu kin t (hoc a f , ( a.b ) ) f f a t t = ữ iu kin hp t>0 b Dng 4: Lng giỏc hoỏ Chỳ ý: Ta s dng ngụn t iu kin hp t>0 cho trng hp t t = a f ( x ) vỡ: - Nu t t = a x thỡ t>0 l iu kin ỳng - Nu t t = x +1 thỡ t>0 ch l iu kin hp, bi thc cht iu kin cho t phi l t iu kin ny c bit quan trng cho lp cỏc bi toỏn cú cha tham s II VD minh ho: - VD1: Gii phng trỡnh: 4cot g x + sin x = (1) Gii: iu kin sin x x k , k Z (*) = + cot g x nờn phng trỡnh (1) c bit di dng: Vỡ sin x cot g x 4cot g x + 2.2 = (2) t t = 2cot g x iu kin t vỡ cot g x 2cot g x 20 = Khi ú phng trỡnh (2) cú dng: t = t + 2t = 2cot g x = cot g x = t = tho (*) cot gx = x = + k , k Z 2 Vy phng trỡnh cú h nghim x = ( ( ) +2=0 Gii: Nhn xột rng: + = ( + ) ; ( + ) ( ) = 1 Do ú nu t t = ( + ) iu kin t>0, thỡ: ( ) = v ( + ) t VD2: Gii phng trỡnh: + ) + k , k Z x 3 x x x x = t2 Khi ú phng trỡnh tng ng vi: t = t + = t + 2t = ( t 1) t + t + = t t + t + = 0(vn) ( ( 2+ ) x ) =1 x = Vy phng trỡnh cú nghim x=0 Nhn xột: Nh vy vớ d trờn bng vic ỏnh giỏ: ( 7+4 = 2+ ) ( + 3) ( 3) =1 Ta ó la chn c n ph t = ( + ) x cho phng trỡnh Vớ d tip theo ta s miờu t vic la chn n ph thụng qua ỏnh giỏ m rng ca a.b=1, ú l: a b a.b = c = tc l vi cỏc phng trỡnh cú dng: A.a x + B.b x + C = c c Khi ú ta thc hin phộp chia c v ca phng trỡnh cho c x , nhn c: x x x x a b a b A ữ + B ữ + C = t ú thit lp n ph t = ữ , t > v suy ữ = c c c c t x +1 x2 + x x+ VD3: Gii phng trỡnh: 9.2 +2 =0 x+ Gii: Chia c v phng trỡnh cho ta c: 2 22 x x 9.2 x x + = 22 x x x x + = x2 x x2 x 2.2 9.2 +4=0 x2 x t t = iu kin t>0 Khi ú phng trỡnh tng ng vi: t = x x = 22 x2 x = x = 2t 9t + = t = x x = x = x x = 21 Vy phng trỡnh cú nghim x=-1, x=2 Chỳ ý: Trong vớ d trờn, vỡ bi toỏn khụng cú tham s nờn ta s dng iu kin cho n ph ch l t>0 v chỳng ta ó thy vi t = vụ nghim Do vy nu bi toỏn cú cha tham s chỳng ta cn xỏc nh iu kin ỳng cho n ph nh sau: 2 1 1 x x = x ữ 2x x 24 t 4 12 3x x VD4: Gii phng trỡnh: 6.2 3( x1) + x = 2 Gii: Vit li phng trỡnh cú dng: x 23 x x ữ x ữ = (1) 23 x 3x = x ữ + 3.2 x x x x 3x 2 ữ = t + 6t x Khi ú phng trỡnh (1) cú dng: t + 6t 6t = t = x = x t u = , u > ú phng trỡnh (2) cú dng: u = 1(1) u u = u2 u = u = 2x = x = u = Vy phng trỡnh cú nghim x=1 Chỳ ý: Tip theo chỳng ta s quan tõm n vic s dng phng phỏp lng giỏc hoỏ t t = x ) ( 2x 2x x VD5: Gii phng trỡnh: + = + 2 Gii: iu kin 22 x 22 x x x Nh vy < x , t = sin t , t 0; ữ Khi ú phng trỡnh cú dng: ) ( + sin t = sin t + sin t + cos t = ( + cos t ) sin t cos t t 3t t t 3t = sin t + sin 2t cos = 2sin cos cos sin ữ = 2 2 2 t x cos = 0(1) t = = x = x 3t x = t = sin = = 2 Vy phng trỡnh cú nghim x=-1, x=0 BI TON 4: S DNG PHNG PHP T N PH- DNG I Phng phỏp: Phng phỏp dựng n ph dng l vic s dng n ph chuyn phng trỡnh ban u thnh phng trỡnh vi n ph nhng cỏc h s cũn cha x Phng phỏp ny thng s dng i vi nhng phng trỡnh la chn n ph cho biu thc thỡ cỏc biu thc cũn li khụng biu din c trit qua n ph ú hoc nu biu din c thỡ cụng thc biu din li quỏ phc Khi ú thng ta c phng trỡnh bc theo n ph ( hoc theo n x) cú bit s l mt s chớnh phng II VD minh ho: 2x x x x VD1: Gii phng trỡnh: + + 9.2 = ( ) Gii: t t = , iu kin t>0 Khi ú phng trỡnh tng ng vi: 2 t = t x + t + 9.2 x = 0; = x + 4.9.2 x = x + x t = Khi ú: + Vi t = 3x = t = x ( ) ( ) ( ) x + Vi t = = ữ = x = Vy phng trỡnh cú nghim x=2, x=0 x2 x2 VD2: Gii phng trỡnh: + x 3 x + = x x x ( ) Gii: t t = 3x iu kin t vỡ x 3x 30 = 2 Khi ú phng trỡnh tng ng vi: t + x t x + = 2 ( ) 2 t = = x x + = x + t = x Khi ú: + Vi t = 3x = x = log x = log ( ) ( ) ( ) + Vi t = x 3x = x ta cú nhn xột: VT VT = 3x = x=0 VP VP = 1 x = Vy phng trỡnh cú nghim x = log 2; x = BI TON 5: S DNG PHNG PHP T N PH- DNG I Phng phỏp: Phng phỏp dựng n ph dng s dng n ph cho biu thc m phng trỡnh v khộo lộo bin i phng trỡnh thnh phng trỡnh tớch II VD minh ho: 2 VD1: Gii phng trỡnh: x x + + x + x + = 42 x +3 x + + 2 2 Gii: Vit li phng trỡnh di dng: x x + + 42 x +6 x +5 = x x + 2.42 x + x + + u = x x + , u, v > t x2 +6 x +5 v = Khi ú phng trỡnh tng ng vi: u + v = uv + ( u 1) ( v ) = x = x = x x + 2 x 3x + = =1 u = x = 42 x +6 x +5 = x + x + v = x = Vy phng trỡnh cú nghim 2 VD2: Cho phng trỡnh: m.2 x x +6 + 21 x = 2.26 x + m(1) a) Gii phng trỡnh vi m=1 b) Tỡm m phng trỡnh cú nghim phõn bit Gii: Vit li phng trỡnh di dng: m.2 x x + 2 + 21 x = 27 x + m m.2 x 2 x + + 21 x = ( ( x x + 6) + x ) +m m.2 x x + + 21 x = x x + 6.21 x + m u = x x + , u , v > Khi ú phng trỡnh tng ng vi: t: x v = x = x x + = u = mu + v = uv + m ( u 1) ( v m ) = x = 21 x = m v = m x2 = m(*) Vy vi mi m phng trỡnh luụn cú nghim x=3, x=2 a) Vi m=1, phng trỡnh (*) cú dng: 21 x = x = x = x = Vy vi m=1, phng trỡnh cú nghim phõn bit: x=3, x=2, x= b) (1) cú nghim phõn bit (*) cú nghim phõn bit khỏc v m > m > (*) Khi ú iu kin l: 2 x = log m x = log m m > m < m > log m > 1 m m ( 0; ) \ ; 256 log m log m m 256 1 Vy vi m ( 0; ) \ ; tho iu kin u bi 256 BI TON 6: S DNG PHNG PHP T N PH- DNG I Phng phỏp: Phng phỏp dựng n ph dng l vic s dng k n ph chuyn phng trỡnh ban u thnh h phng trỡnh vi k n ph Trong h mi thỡ k-1 thỡ phng trỡnh nhn c t cỏc mi liờn h gia cỏc i lng tng ng Trng hp c bit l vic s dng n ph chuyn phng trỡnh ban u thnh h phng trỡnh vi n ph v n x, ú ta thc hin theo cỏc bc: Bc 1: t iu kin cú ngha cho cỏc biu tng phng trỡnh Bc 2: Bin i phng trỡnh v dng: f x, ( x ) = y = ( x ) Bc 3: t y = ( x ) ta bin i phng trỡnh thnh h: f ( x; y ) = II VD minh ho: 2x 18 VD1: Gii phng trỡnh: x + x = x 1 x +1 + 2 + + 18 + x = x 1 x Gii: Vit li phng trỡnh di dng: x +1 +1 + + x u = + , u, v > t: x v = + ( )( ) x 1 x x 1 x Nhn xột rng: u.v = + + = + + = u + v Phng trỡnh tng ng vi h: 18 u = v = u + 8v = 18 + = u v u + v u = 9; v = u + v = uv u + v = uv x + = x =1 + Vi u=v=2, ta c: x + = x + = 9 + Vi u=9 v v = , ta c: x x=4 + = Vy phng trỡnh ó cho cú cỏc nghim x=1 v x=4 VD2: Gii phng trỡnh: 22 x x + = Gii: t u = x , iu kin u>0 Khi ú phng trỡnh thnh: u u + = t v = u + 6, iu kin v v = u + Khi ú phng trỡnh c chuyn thnh h: u v = u = v + u v2 = ( u v ) ( u v ) ( u + v ) = v = u + u + v + = u = 2x = x = + Vi u=v ta c: u u = u = 2(1) + Vi u+v+1=0 ta c: + 21 u = 21 21 u2 + u = 2x = x = log 2 21 (1) u = 21 Vy phng trỡnh cú nghim l x=8 v x= log BI 7: S DNG TNH CHT N IU CA HM Sễ I Phng phỏp: S dng cỏc tớnh cht ca hm s gii phng trỡnh l dng toỏn khỏ quen thuc Ta cú hng ỏp dng: Hng1: Thc hin cỏc bc sau: Bc 1: Chuyn phng trỡnh v dng: f(x)=k Bc 2: Xột hm s y=f(x) Dựng lp lun khng nh hm s n iu( gi s ng bin) Bc 3: Nhn xột: + Vi x = x0 f ( x ) = f ( x0 ) = k ú x = x0 l nghim + Vi x > x0 f ( x ) > f ( x ) = k ú phng trỡnh vụ nghim + Vi x < x0 f ( x ) < f ( x0 ) = k ú phng trỡnh vụ nghim Vy x = x0 l nghim nht ca phng trỡnh Hng 2: Thc hin theo cỏc bc: Bc 1: Chuyn phng trỡnh v dng: f(x)=g(x) Bc 2: Xột hm s y=f(x) v y=g(x) Dựng lp lun khng nh hm s y=f(x) l L ng bin cũn hm s y=g(x) l hm hng hoc nghch bin Xỏc nh x0 cho f ( x0 ) = g ( x0 ) Bc 3: Vy phng trỡnh cú nghim nht x = x0 Hng 3: Thc hin theo cỏc bc: Bc 1: Chuyn phng trỡnh v dng: f(u)=f(v) (3) Bc 2: Xột hm s y=f(x) Dựng lp lun khng nh hm s n iu ( gi s ng bin) Bc 3: Khi ú: (3) u = v vi u , v D f II VD minh ho: VD1: Gii phng trỡnh: x + 2.3log x = (1) Gii: iu kin x>0 Bin i phng trỡnh v dng: 2.3log x = x (2) Nhn xột rng: + V phi ca phng trỡnh l mt hm nghch bin + V trỏi ca phng trỡnh l mt hm ng bin Do vy nu phng trỡnh cú nghim thỡ nghim ú l nht Nhn xột rng x=1 l nghim ca phng t rỡnh (2) vỡ 2.3log2 x = Vy x=1 l nghim nht ca phng trỡnh ( x x ) VD2: Gii phng trỡnh: log x x + + + ữ x Gii: iu kin: x 3x + x 2 = (1) t u = x 3x + , iu kin u suy ra: x 3x + = u 3x x = u u Khi ú (1) cú dng: log ( u + ) + ữ x Xột hm s: f ( x) = log ( x + ) + ữ + Min xỏc nh D = [ 0; +) =2 = log ( x + ) + x 1 + x.5 x ln > 0, x D Suy hm s tng trờn D ( x + ) ln Mt khỏc f ( 1) = log ( + ) + = Do ú, phng trỡnh (2) c vit di dng: f ( u ) = f ( 1) u = x x + = x = Vy phng trỡnh cú hai nghim x = 2 x +4 mx +2 VD2: Cho phng trỡnh: x + mx + = x + 2mx + m a) Gii phng trỡnh vi m = b) Gii v bin lun phng trỡnh Gii: t t = x + 2mx + phng trỡnh cú dng: 5t + t = 52t + m + 2t + m (1) t Xỏc nh hm s f ( t ) = + t + o hm: f = + Min xỏc nh D=R + o hm: f = 5t.ln + > 0, x D hm s tng trờn D Vy (1) f ( t ) = f ( 2t + m ) t = 2t + m t + m = x + 2mx + m = (2) x = 4 2 a) Vi m = ta c: x + x = x x = x = 5 5 Vy vi m = phng trỡnh cú 2nghim x = 2; x = 5 b) Xột phng trỡnh (2) ta cú: ' = m m + Nu ' < m m < < m < Phng trỡnh (2) vụ nghim phng trỡnh (1) vụ nghim + Nu ' = m=0 hoc m=1 vi m=0 phng trỡnh cú nghim kộp x=0 vi m=1 phng trỡnh cú nghim kộp x0=-1 m > + Nu ' > phng trỡnh (2) cú nghim phõn bit x1,2 = m m m ú cng l m < nghim kộp ca (1) Kt lun: Vi m=0 phng trỡnh cú nghim kộp x=0 Vi m=1 phng trỡnh cú nghim kộp x0=-1 Vi 01 nờn s bin thiờn ca hm s ph thuc vo s bin thiờn cca hm s t = x x + ta cú: a) Vi m=8 phng trỡnh cú nghim nht x=1 b) Vi m=27 phng trỡnh cú nghim phõn bit x=0 v x=2 c) Phng trỡnh cú nghim m>8 x2 x +3 VD2: Vi giỏ tr no ca m thỡ phng trỡnh: ữ = m m2 + cú nghim phõn bit Gii: Vỡ m m + > vi mi m ú phng trỡnh tng ng vi: x x + = log m m + ( ) ( ) 2 t log m m + = a , ú: x x + = a Phng trỡnh ban u cú nghim phõn bit phng trỡnh (1) cú nghim phõn bit ng thng y=a ct th hm s y = x x + ti im phõn bit x x + 3khix 1hoacx y = x x + = Xột hm s: x x + 3khi1 x x 4khix < 1hoacx > o hm: y ' = x + 4khi1 < x < Bng bin thiờn: 10 2 2 x3 x3 x3 x3 log ữ = log ữ = log ữ = log ữ = log x log 2 log ( x ) = log ( x ) = log ( x ) = log 2 ( x ) BI TON 4: S DNG PHNG PHP T N PH- DNG I Phng phỏp: II VD minh ho: Gii bt phng trỡnh: log x log ( x ) log x + log x < (1) Gii: iu kin x>0 Bin i phng trỡnh tng ng v dng: log x ( + log x ) log x + 3log x < t t = log x ú bt phng trỡnh cú dng: f ( t ) = t ( + log x ) t + 3log x < (2) t = 2 Ta cú: = ( + log x ) 12 log x = ( log x ) Do ú f(t)=0 cú nghim: t = log x Do ú (2) tng ng vi: ( t 3) ( t log x ) < ( log x 3) ( log x log x ) < log x > log x > x > 27 x > 27 log x log x < log x < log x x > x < 27 log x < log x < < x < log x log x > log x > log x < x < Vy bt phng trỡnh cú nghim l ( 0;1) ( 27; + ) BI TON 5: S DNG PHNG PHP T N PH- DNG I Phng phỏp: S dng n ph cho biu thc m bt phng trỡnh v bin i bt phng trỡnh thnh bt phng trỡnh tớch, ú lu ý: A > A > B > B < A.B > A.B < v A < A < B < B > II VD minh ho: x Gii bt phng trỡnh: log x.log x < log x log Gii: iu kin x>0 (*) Vit li bt phng trỡnh di dng: log x.log x log x log x < u = log x t Khi ú bt phng trỡnh cú dng: v = log x 39 uv 2u v < ( u 1) ( v ) < log x > u > x > tho (*) v < log x < x < 3< x < u < x < log x < log x > v > x > Vy bt phng trỡnh cú nghim 3 VD2: Gii bt phng trỡnh: log x x + log ( x + 1) 3 x > < x < x < < x < 2 < x x + Gii: iu kin: x < x < < x + x x > < x 2 Ta cú: A = log x x + > x x + < 3 B = log ( x + 1) > x + < x < x 3x + < < x < T ú ta cú bng xột du sau: 40 + Vi -1 ( x + 1) Kt hp vi trng hp ang xột ta c x>5 Vy bt phng trỡnh cú nghim: 0; ữ 1; ữ ( 5; + ) CH 3: H PHNG TRèNH LễGARIT BI TON 1: S DNG PHNG PHP BIN I TNG I Phng phỏp: Ta thc hin theo cỏc bc sau: Bc 1: t iu kin cho cỏc biu thc h cú ngha Bc 2: S dng cỏc phộp th nhn c t h phng trỡnh theo n x hoc y (ụi cú th l theo c n x, y) Bc 3: Gii phng trỡnh nhn c bng cỏc phng phỏp ó bit i vi phng trỡnh cha cn thc Bc 4: Kt lun v nghim cho h phng trỡnh II VD minh ho: x (1) x + 3y = VD1: Gii h phng trỡnh: x y + log x = 1(2) ( ) x +1 Gii: iu kin: x < x x > log x y T phng trỡnh (2) ta c: y = log x = 3 = log3 x = (3) x Th (3) vo (1) ta c: 3 x x +1 = x +1 = x x + = x + x x x x x = x2 x = y = x 3x = x = ( x ) Vy h phng trỡnh cú cp nghim (3;0) x y = VD2: Gii h phng trỡnh: log ( x + y ) log ( x y ) = x + y > Gii: iu kin: (*) x y > ( ) 41 T phng trỡnh th nht ca h ly lụgarit c s hai v ta c: log x y = log 2 log ( x + y ) + log ( x y ) = ( ) log ( x + y ) = log ( x y ) Th vo phng trỡnh th hai ta c: log ( x y ) log 2.log ( x y ) = ( + log ) log ( x y ) = log ( x y ) = x y = x= x + y = x y = Vy ta c h mi: tho iu kin (*) x y = x y = y = Vy h phng trỡnh cú nghim BI TON 2: S DNG PHNG PHP T N PH I Phng phỏp: Phng phỏp c s dng nhiu nht gii cỏc h lụgarit l vic s dng cỏc n ph Tu theo dng ca h m la chn phộp t n ph thớch hp Ta thc hin theo cỏc bc sau: Bc 1: t iu kin cho cỏc biu thc ca h cú ngha Bc 2: La chn n ph bin i h ban u v cỏc h i s ó bit cỏch gii (h i xng loi I, loi II v h ng cp bc hai) Bc 3: Gii h nhn c Bc 4: Kt lun v nghim cho h ban u II VD minh ho: xy + yx = 32 Gii h phng trỡnh: log ( x y ) = log ( x + y ) x y > Gii: iu kin: x + y > x; y x y x y + ữ = + ữ = 5(1) y x Bin i h phng trỡnh v dng: y x log x y = 2 x y = 3(2) ( ) x y = Khi ú (1) cú dng: y x t t = x = 2y 2 t + ữ = 2t 5t + = t = t y = 2x y =1 x = 2 + Vi x=2y (2) y y = y = x = 2(1) Gii (1): t t = + Vi y=2x (2) x y = vụ nghim Vy h phng trỡnh cú cp nghim (2;1) 42 BI TON 3: S DNG PHNG PHP HM S I Phng phỏp Ta thc hin theo cỏc bc sau: Bc 1: t iu kin cho biu thc ca h cú ngha Bc 2: T h ban u chỳng ta xỏc nh c phng trỡnh h qu theo n hoc theo c n, gii phng trỡnh ny bng phng phỏp hm s ó bit Bc 3: Gii h mi nhn c II VD minh ho: log x + = + log y Gii h phng trỡnh: log y + = + log x Gii: iu kin x; y>0 Bin i tng ng h v dng: log ( x + 3) = ( + log y ) log ( x + ) = ( + log y ) log ( y + 3) = ( + log x ) ( + log x ) = log ( y + ) (I) log ( x + 3) + log3 x = log ( y + 3) + log y (1) Xột hm s: f ( t ) = log ( t + 3) + log t Min xỏc nh D = ( 0; + ) + > 0, t D hm s luụn ng bin o hm f ( t ) = ( t + 3) ln t.ln Vy phng trỡnh (1) c vit di dng: f ( x ) = f ( y ) x = y x = y Khi ú h (I) tr thmh: (II) log ( x + 3) = ( + log x ) (2) 2 + Gii (2): x + = 22( 1+ log3 x ) x + = 4.2log3 x x + = 4.2log3 2.log2 x x + = ( x ) log3 x + = 4.x log3 x1log3 + 3.x log3 = (3) log log Xột hm s g ( x ) = x + 3.x Min xỏc nh D = ( 0; + ) log log < 0x D hm s luụn nghch bin o hm: g ' ( x ) = ( log ) x 3log 4.x Vy phng trỡnh (3) nu cú nghim thỡ nghim ú l nht Nhn xột rng nu x=1 l nghim ca phng trỡnh bi ú: 11log3 + 3.11log3 = = ỳng x = y x = y =1 Khi ú h (II) tr thnh: x = Vy h ó cho cú nghim nht (1;1) BI TON 4: S DNG PHNG PHP NH GI I Phng phỏp: II VD minh ho: e x e y = ( log y log x ) ( xy + 1) (1) VD1: Gii h phng trỡnh: 2 x + y 1(2) Gii: iu kin x; y>0 *) Gii (1) ta cú nhn xột sau: 43 VT( 1) - Nu x > y log x > log y , ú: VP( 1) VT( 1) - Nu x < y log x < log y , ú: VP( 1) - Vy x=y l nghim ca (1) >0 x + y > Gii: iu kin: xy + > 0 < x + y + xy + > T phng trỡnh th nht ca h vi vic s dng n ph t=x+y>0, ta c: log t = t u t u = log t t = ú phng trỡnh cú dng: log t = u = x + y = Bernoulli 2u = u + u = x + y = log t = x + y = x + y = x + y = x = 0; y = + Vi x+y=1 h cú dng: xy + = xy = x = 1; y = log ( xy + 1) = x + y = x + y = x + y = + Vi x+y=2 h cú dng: xy = log ( xy + 1) = xy + = Khi ú x; y l nghim ca phng trỡnh: t 2t + = vụ nghim Vy h cú cp nghim (0;1) v (1;0) 44 PHNG TRèNH M x +1 x+4 x+2 1) + = + 2) x +8 4.3 x +5 + 27 = 30) 31) 32) 33) 34) 35) 36) 37) 38) x 3) 4.3 x 9.2 x = 5.6 4) 8.3 x + 3.2 x = 24 + x 72x x 5) = 6.( 0.7 ) + x 100 6) 125 x + 50 x = x +1 7) x + x.3 x + 31+ x = x x + x + x 8) x.8 x = 500 9) x +1 + x x + x = 750 10) 7.3 x +1 x + = x + x +3 11) 6.4 x 13.6 x + 6.9 x = 12) x = x 13) x +1 3.5 x = 110 14) 3.4 x + 2.9 x = 5.6 x 15) x +8 4.3 x +5 + 27 = 16) 7.3 x +1 x + = x + x +3 x 17) 6.9 13.6 ( 18) + x ) + x x x +1 ( x ( ) x+2 ) ( ) ( )( x 28) + = x 29) x = 128 ( 44) 3.16 + 2.81 = 2.36 ( x 45) + 101 = 10 ( =0 23) + + + 24) 25 x + 15 x = 2.9 x 25) x + 16 = 10.2 x 2 26) 2 x +1 9.2 x + x + 2 x + = 12 3x x 27) 6.2 3( x 1) + x = 2 ) x x x 21) x x + = x + 32 x 22) x + log x = x ( + 6.4 = 19) + + 2 20) 2 x = x +3 x x 39) x + + x = x +1 40) x = 32 x + 2.5 x + 2.3 x 2 2 41) x x = x x + 2 x x x 42) = 10 x x 43) + + 16 = x +3 x + 4x + 2x = 25 X 6.5 x +1 + 53 = x + 5.3 x + = x 25.3 x 54 = 2+ x + 32 x = 30 2( x +1) 82.3 x + = x + 9.5 x = x + 9.7 x 2 x 36.3 x + = 2 x +1 x +1 = ) x ( = 2+ ) ) ( ) lo2 x ) x + x log x = 1+ x2 ) 46) x x + x = x + x 47) x log2 = x 3log2 x x log2 x 48) x.8 x + = 49) 2.x log2 x + x log8 x = 50) x + x log = x log 51) ( x 2) log 4( x ) = 4( x ) 52) lg10 x lg x = 2.3lg100 x x x x 1 53) + x = x + 6 54) 5.3 x 7.3 x + 6.3 x + x +1 = 55) 12.3 x + 3.15 x x +1 = 20 56) log2 x x log2 = 2.3log2 x 57) x + x = x + 2 58) x x x = ( x 1) x PHNG TRèNH LễGARIT 45 1) log ( x + 3) log ( x 1) = log 2) log x x x + 65 = 3) lg + lg( x + 10 ) = lg( 21x 20 ) lg( x 1) 1 4) lg x lg x = lg x + lg x + 2 2 5) lg x lg x = lg x ( ) x2 =8 7) log ( x ) + log 8) log (4 x ) ( ) log x = x + log x x = log x x 9) log x log 2 + log ( x ) = x 45) log ( ) x2 +1 x = ( ) x + log x + log 3 x = 3 ) ( ) 50) log x + log x = + log x log x 51) ) ( log x x log x + x = log 20 x x 11) log x x + 40 log x x 14 log16 x x = ( ) [ ] [log (9 6)] = ( + 4) x = log ( 52) log x + 5.3 x = x +1 x 53) log x 4.3 = x + 12) log x.log x.log x = log x.log x + log x.log x + log x.log x x ) ( ( x = + log x 13) log ( x ) log x log 3 14) lg( lg x ) + lg lg x = 15) log ( x + 1) + log ( x + 1) = 16) log x log ( x 10) + = [ ( ) ] ( ) x + + x + log ( 10) log log x = ) ( ) x + x = log x x x +1 49) log + = x log 3 ( 44) 46) log x log x + 2 43) log 2+ ( =0 47) log x + log x = log log 225 = + log ( x + 1) 48) log ( x 1) + log x +3 6) log x log x + = 42) log ) 17) log x + x = log x 18) log x x x = 54) log x x 55) log x ( ) log x + = log 16 56) x +1 ( ) ( log x ) x log x 57) + 2 + x 2 58) log x = log ( x 1) ( ) ) + 12 = 1+ x2 59) log ( log ( log x ) ) = 60) 61) x lg x + + x lg x + = ( ) ( ) ( ) ( ) 19) log x x log x = 12 20) log x ( x + 1) lg 4,5 = 62) log ( x + 1) + ( x 5) log ( x + 1) x + = 63) 3 21) log x + + log x = x x 22) x + lg x x = + lg( x + ) 23) log x log ( x 10) + = log x x log x + x = log x x ( 24) log 25) log 2+ ( 2 ) (x ) x = log 2+ 26) log x = log x log 64) log x + log x + log x = ) x log x 3 + log [ ) 2x ] 2x +1 27) log x log 3 x = ( ) 16 ( ) ( ) x +1 65) log 5 log 25 = 66) log x + log x = log log 225 67) log ( x + 8) log ( x + 26 ) + = 3= (x ) ( ( x 68) x log x 27 log x = x + 69) log ( x + ) + log x + x + = 70) log12 x x x + log13 x x x + = ( 46 ) ( ) ) ) ( 28) x log 2 x + log x x + log 4 29) 30) 31) 32) 33) = log x lg( x + 10) + lg x = lg log ( x +1) 3x = log x + log x ( x + 3) log 32 ( x + 2) + 4( x + 2) log ( x + 2) = 16 log ( x +3 ) x + x = 2 log 36 + log 81 = log x x 15 log ( ) ) ( ( ) 34) log x +1 x + = 35) 36) log 22 x + ( x 1) log x + x = 37) log x + log x = ( ) 38) log x = log x 39) log x + x + + log x + x + 12 = + log ( 40) ) ( ) 72) log ( x + 1) + log ( ) ( ) 71) x lg x + + x lg x + = ( ) ( ) ( ) ) x + + x log x x = ( ) ( = log ( x + 2) log ( x ) 5 25 73) ( x + ) log ( x + 1) + 4( x + 1) log ( x + 1) 16 = 3 74) log ( x + ) = log ( x ) + log ( x + ) 4 3 x3 = + log x 75) log log x log x 76) log x +7 + 12 x + x + log x +5 x + 23 x + 21 = 77) ( ) ( ) ( ) x log x x x log x x = x + x 78) log x log x = log x + log x 3 ( x 3) log ( x 1) + log x 79) = ( x 3) log x + log x x 80) log x + x log ( x + 3) = + log ( x + 3) log x ) x x +1 41) log 5 log 25 = BT PHNG TRèNH M 1) + x x 15) x + x x + 31+ +1 x x 2) + = 12 3) 16 loga x + 3.x loga 4) ( ) +1 x2 + x + 2x 16) 9) 2 x +1 21 2 ( x 1) + x +1 ( ) < x2 + x 19) log3 21) x 13) 6.9 z x 13.6 x +20 x + 6.4 x ( x ) x x +1 x x +1 52 ) < 2.3 x x + x + 17) 25 +9 34.15 x x 18) ( log5 x ) + x log5 x 10 ) 8) x 2( x 1) + 5+2 x x +1 5) x 8.3 x + x + 9.9 x + > 6) x + x x 7) x +1 16 x < log ( ( x x x 28) x +1 2 x +1 12 < 47 14) x x + x > x.3 x x 3x + x x BT PHNG TRèNH LễGARIT x + 8x x +1 x x 2) log 2 + + log + 2 3) log ( (x 4) ) ) ( ) x + ( ) ( log 3x + x + + > log 3x + x + x ) 6x 2x 1 < 6) log x 7) log x x 32 x 8) log x log + log < log x x 2 5) log x3 ( ) 10) log x ( x + 5) 11) log x + log x 12) x x x + + log + 8x x + x 13) x 16 x + log ( x 3) 14) log x x + + log x > log ( x 3) 3 ) ( ) ( ) 2x 17) log x x + log ( x + 1) 15) log 3 ( ( 20) log ( x 21) ) ( ) ( ) 57) log x 64 + log x 16 2 58) log x + log x < 59) log x x 60) x + log x x + > ( x + 1) log ( x ) ( 61) log ( ) x+ x ) log x +1 log ( x 1) 62) log 25 ( x 1) log 2x 1 ) ) 63) 18) log x + x 2 19) log x 11x + 43 < 2 2x >1 48) log x x + log 32 x >1 49) + log x 50) log x log x > log 35 x >3 51) log ( x ) 52) log x x +1 x x < 53) log x log x > 2 log ( x + 1) log ( x + 1) 54) >0 x 3x lg x x + >2 55) lg x + lg 2 56) log x x 18 x _ + 16 > ( 9) x + log x x + > ( x + 1) log ( x ) ( log x 47) log x 2( + log x ) > 1) log ( ) ( log x + x + + > log 2 x + x + 32 x + log < log 21 x log x log 64) x 2 ) x + < 65) log ( x + 1) log ( x ) ( ( ) ( x x 66) log + > log + 2 48 ) log 22 x + log x > log x ) ) ( x + 6x + < log ( x + 1) 2( x + 1) 22) log b log3 x log3 x < 18 x 23) log 18 log x 24) log x log < log ( x 1) + log ( x + 1) + log ( x ) < 25) ( x [ ) 3 ) ( ) d log x 6x + + log5 ( x ) < 5 e log x + log x 26) log x x ( x ) > ( ( c log log x > )] ( ) a log8 x 4x + ) 27) log x x + x > x 28) f log x log < 2 2 + x x + 12 14 x x 24 + log x g log x 2.log 2x 2.log 4x > x x 4x + 29) log x x x + > h log x 30) = i log2 ( x + ) + log ( x 1) 3x x log log 31) 16 4 j log8 (x 2) + log (x 3) > 32) log 0,3 x + x + > log log x ( ) log x x + + log x > log x + k ữ 33) 1 ữ 3 l log5 3x + 4.log x > log x x + 11 log11 x + 11 >0 34) x 4x + 2 x 3x m log3 x2 + x 35) ( log x ) log x log x + log3 x > n 4x o log2x x 5x + < 36) log x x 2 p log3x x2 ( x ) > 1 37) log ( x 1) > log 1 x 2 log 3x x x + ữ q x2 +1 38) log x log x > 2 x >0 r log x+6 log2 x5 x+2ữ 39) log ( x ) s log2 x + log x 40) log x + < log ( x ) 1 + log ( x + 2) t log x 2.log x > log x > 41) 16 2x +1 x log ( x 1) u log32 x log3 x + log3 x 42) log x ) 49 ( ) [ ( )] x 44) log x log 45) log x log x < log x log x ( ) log x x + log x (D206) 2(log x + 1) log x + log = 51 (3 < x3 ) (x>2 x < ) ( x=2 x= ẳ) 4) 5) 6) log x + 2log 0,25 ( x 1) + log (B203) 0,5 log x - + log x + + log = ổ ỗ ỗ ỗx ẻ ỗ ố ỡù ỹử ữ ùớ - 6;3; - 17 ùùýữ ữ ữ ùù ùù ứ ợ ỵữ ổ 17 ữ ỗ ữ ỗ x = 6; x = ữ ỗ ữ ỗ ứ ố log ( x + 2) + log ( x - 5) + log = 8) 1 log ( x + 3) + log ( x 1)8 = log (4 x) log ( x + 3) - log x - - log < 9) 7) (x 3) (x = x= 3+ 12 ) (- 4; - 3) ẩ (- 3; - 1) ẩẩ(0; 2) (2;3) log x+ ( -10 < x < ) 5log x+ < 400 x + x 16 10) (B104) >4 x2 3 (x 4) 11) (A104) log [log ( x + x x )] < (x >1 x< - 4) 12) (B204) log x > log x 2 13) (D03) x x 22 + x x = 14) (D2.05) x 2 x 2. ( x>3 1/3