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8/7/2019 phuong phap giai phuong trinh
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CC PHNG PHP GIIPHNG TRNH- BT PHNG TRNH- H M- LGARIT
CHNG I:PHNG PHP GII PHNG TRNH- BT PHNG TRNH- H MCH I:PHNG TRNH M
BI TON 1: S DNG PHNG PHP BIN I TNG NGI. Phng php:
Ta s dng php bin i tng ng sau:
( ) ( )
( ) ( )
1
0 1f x g x
a
aa a
f x g x
= < = =
hoc ( ) ( ) ( )0
1 0
a
a f x g x
> =
II. VD minh ho:
VD1: Gii phng trnh: ( ) ( )sin 2 3 cos2 22 2 xx x x x + = + Gii: Phng trnh c bin i v dng:
( ) ( )
2
2
2
1 2(*)2 0
1 0(1)
2 1 sin 2 3 cos 0 sin 3 cos 2(2)
xx x
x x
x x x x x x
< =
+ + = + =
Gii (1) ta c1,2
1 5
2x
= tho mn iu kin (*)
Gii (2):1 3
sin cos 1 sin 1 2 2 ,2 2 3 3 2 6
x x x x x k x k k Z
+ = + = + = + = +
nghim tho mn iu kin (*) ta phi c:1 1
1 2 2 1 2 0,6 2 6 2 6
k k k k Z
< + < < < =
khi ta nhn c 3 6x
=
Vy phng trnh c 3 nghim phn bit 1,2 31 5 ;2 6x x = = .
VD2: Gii phng trnh: ( ) ( )22 43 5 2 23 6 9
x xx xx x x
+ + = +
Gii: Phng trnh c bin i v dng: ( ) ( ) ( )2
2 243 5 2 2 2( 4)3 3 3
x xx x x x
x x x+ + + = =
2 2 2
3 1 44
0 3 1 3 45
3 5 2 2 2 8 7 10 0
x xx
x xx
x x x x x x
= = = < < = + = + + =
Vy phng trnh c 2 nghim phn bit x=4, x=5.
BI TON 2: S DNG PHNG PHP LGARIT HO V A V CNG C SI. Phng php: chuyn n s khi s m lu tha ngi ta c th logarit theo cng 1 c s c 2 v ca phngtrnh, ta c cc dng:Dng 1: Phng trnh:
( )
( )0 1, 0
logf x
a
a ba b
f x b
< >= =Dng 2: Phng trnh :
1
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( ) ( ) ( ) ( )log log ( ) ( ).logf x g x f x f xa a aa b a b f x g x b= = =
hoc ( ) ( )log log ( ).log ( ).f x g xb b ba b f x a g x= =II. VD minh ho:VD1: Gii phng trnh:
22 2 3
2x x =
Gii: Ly logarit cs 2 hai v phng trnh ta c:
2 2 2 22 2 2 2
3log 2 log 2 log 3 1 2 1 log 3 0
2x x x x x x = = + =
Ta c , 2 21 1 log 3 log 3 0 = + = > suy ra phng trnh c nghim
x = 1 2log 3. VD2: Gii phng trnh:
1
5 .8 500.x
x x
=Gii: Vit li phng trnh di dng:
1 1 33 3 2 385 .8 500 5 .2 5 .2 5 .2 1
x x xx x xx x
= = =
Ly logarit c s 2 v, ta c:
( ) ( )3 3
3 32 2 2 2 2
3log 5 .2 0 log 5 log 2 0 3 .log 5 log 2 0
x x
x xx xx
xx
= + = + =
( ) 22
31
3 log 5 0 1
log 5
x
xxx
= + = =
Vy phng trnh c 2 nghim phn bit:2
13;
log 5x x= =
Ch : i vi 1 phng trnh cn thit rt gn trc khi logarit ho.BI TON 3: S DNG PHNG PHP T N PH- DNG 1I. Phng php:Phng php dng n ph dng 1 l vic s dng 1 n ph chuyn phng trnh ban u thnh 1phng trnh vi 1 n ph.Ta lu cc php t n ph thng gp sau:Dng 1: Phng trnh ( 1)1 1 0..... 0
k x x
k ka a + + =
Khi t xt a= iu kin t>0, ta c: 11 1 0...... 0k k
k kt t t + + =
M rng: Nu t ( ) ,f xt a= iu kin hp t>0. Khi : 2 ( ) 2 3 ( ) 3 ( ), ,.....,f x f x kf x k a t a t a t = = =
V( ) 1f x
a t
=Dng 2: Phng trnh 1 2 3 0
x xa a + + = vi a.b=1
Khi t ,xt a= iu kin t0, suy ra ( )1f x
bt
=
2
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Dng 3: Phng trnh ( )2 21 2 3 0xx xa ab b + + = khi chia 2 v ca phng trnh cho 2xb >0
( hoc ( )2 , . xxa a b ), ta c:2
1 2 3 0x x
a a
b b
+ + =
t ,x
at
b
=
iu kin t (hoc ( )2 , . ffa a b )
- tf
at
b
=
iu kin hp t>0
Dng 4: Lng gic ho.Ch : Ta s dng ngn t iu kin hp t>0 cho trng hp t ( )f xt a= v:
- Nu t xt a= th t>0 l iu kin ng.- Nu t
2 12xt += th t>0 ch l iu kin hp, bi thc cht iu kin cho t phi l 2t . iukin ny c bit quan trng cho lp cc bi ton c cha tham s.
II. VD minh ho:VD1: Gii phng trnh: 2 2
1
cot sin4 2 3 0g x x+ = (1)
Gii: iu kin sin 0 ,x x k k Z (*)
V 221
1 cotsin
g xx
= + nn phng trnh (1) c bit di dng:
22 cotcot4 2.2 3 0
g xg x + = (2)
t2cot2 g xt= iu kin 1t v
22 cot 0cot 0 2 2 1g xg x =Khi phng trnh (2) c dng:
22 cot 21
2 3 0 2 1 cot 03
cot 0 ,2
g xt
t t g xt
gx x k k Z
=+ = = = =
= = + tho mn (*)
Vy phng trnh c 1 h nghim ,2
x k k Z
= +
VD2: Gii phng trnh: ( ) ( )7 4 3 3 2 3 2 0x x
+ + =
Gii: Nhn xt rng: ( ) ( ) ( )27 4 3 2 3 ; 2 3 2 3 1+ = + + =Do nu t ( )2 3
x
t= + iu kin t>0, th: ( ) 12 3x
t = v ( ) 27 4 3
x
t+ =
Khi phng trnh tng ng vi:
( ) ( )2 3 2 213
2 0 2 3 0 1 3 03 0 ( )
tt t t t t t
t t t vn
= + = + = + + = + + =
( )2 3 1 0x x + = =Vy phng trnh c nghim x=0Nhn xt: Nh vy trong v d trn bng vic nh gi:
3
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( )
( ) ( )
2
7 4 3 2 3
2 3 2 3 1
+ = +
+ =
Ta la chn c n ph ( )2 3 xt= + cho phng trnhV d tip theo ta s miu t vic la chn n ph thng qua nh gi m rng ca a.b=1, l:
. . 1a b
a b c c c= = tc l vi cc phng trnh c dng: . . 0x xA a B b C + + =
Khi ta thc hin php chia c 2 v ca phng trnh cho 0xc , nhn c:
. 0
x xa b
A B Cc c
+ + =
t thit lp n ph , 0x
at t
c
= >
v suy ra1
xb
c t
=
VD3: Gii phng trnh:2 22 1 2 22 9.2 2 0x x x x+ + + + =
Gii: Chia c 2 v phng trnh cho 2 22 0x+ ta c:
2 2 2 22 2 1 2 2 2 21 92 9.2 1 0 .2 .2 1 0
2 4
x x x x x x x x + = + =
2 22 22.2 9.2 4 0x x x x + =
t2
2x xt = iu kin t>0. Khi phng trnh tng ng vi:2
2
2 22
21
42 2 2 1
2 9 4 0 1212 2
2
x x
x x
tx x x
t txt x x
= = = = + = == = = Vy phng trnh c 2 nghim x=-1, x=2.Ch : Trong v d trn, v bi ton khng c tham s nn ta s dng iu kin cho n ph ch l t>0
v chng ta thy vi1
2t= v nghim. Do vy nu bi ton c cha tham s chng ta cn xc nh
iu kin ng cho n ph nh sau:
2
2 1
2 4
4
1 1 1 12 22 4 4 2
x xx x x t =
VD4: Gii phng trnh: ( )3
3 1
1 122 6.2 1
22
x x
xx + =
Gii: Vit li phng trnh c dng:
3
33
2 22 6 2 1
2 2x x
x x
=
(1)
t33
3 33
2 2 2 22 2 2 3.2 2 6
2 2 2 2x x x x x
x x x xt t t
= = + = +
Khi phng trnh (1) c dng: 3 26 6 1 1 2 12
x
xt t t t + = = =
t 2 , 0xu u= > khi phng trnh (2) c dng:
21(1)
1 2 0 2 2 2 122
xuu
u u u u xu
= = = = = = =
Vy phng trnh c nghim x=1Ch : Tip theo chng ta s quan tm n vic s dng phng php lng gic ho.
VD5: Gii phng trnh: ( )2 21 1 2 1 2 1 2 .2x x x+ = +
4
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Gii: iu kin 2 21 2 0 2 1 0x x x
Nh vy 0 2 1x< , t 2 sin , 0;2
x t t =
Khi phng trnh c dng:
( ) ( )2 21 1 sin sin 1 2 1 sin 1 cos 1 2 cos sin3 32 cos sin sin 2 2 cos 2sin cos 2 cos 1 2 sin 0
2 2 2 2 2 2
cos 0(1) 12 12 6
203 2
2 1sin22 2
x
x
t t t t t t
t t t t t t t t
tt
x
xtt
+ = + + = +
= + = = = = = = = = ==
Vy phng trnh c 2 nghim x=-1, x=0.BI TON 4: S DNG PHNG PHP T N PH- DNG 2I. Phng php:Phng php dng n ph dng 2 l vic s dng 1 n ph chuyn phng trnh ban u thnh 1
phng trnh vi 1 n ph nhng cc h s vn cn cha x.Phng php ny thng s dng i vi nhng phng trnh khi la chn n ph cho 1 biu thc thcc biu thc cn li khng biu din c trit qua n ph hoc nu biu din c th cngthc biu din li qu phc tp.Khi thng ta c 1 phng trnh bc 2 theo n ph ( hoc vn theo n x) c bit s l mt schnh phng.II. VD minh ho:
VD1: Gii phng trnh: ( )23 2 9 .3 9.2 0x x x x + + =Gii: t 3xt= , iu kin t>0. Khi phng trnh tng ng vi:
( ) ( ) ( )
2 229
2 9 9.2 0; 2 9 4.9.2 2 92
x x x x x
x
tt t
t
= + + = = + = +
=
Khi :+ Vi 9 3 9 2xt t= = =
+ Vi3
2 3 2 1 02
x
x x xt x = = = =
Vy phng trnh c 2 nghim x=2, x=0.
VD2: Gii phng trnh: ( )2 22 29 3 3 2 2 0x xx x+ + =Gii: t
2
3xt= iu kin 1t v22 00 3 3 1xx =
Khi phng trnh tng ng vi:
( )2 2 23 2 2 0t x t x+ + =
( ) ( ) ( )2 22 2 2 22
3 4 2 2 11
tx x x
t x
= = + = + =
Khi :
+ Vi2 2
3 32 3 2 log 2 log 2xt x x= = = =
+ Vi22 21 3 1xt x x= = ta c nhn xt:
2
2
1 1 3 10
1 1 1 1
xVT VT x
VP VP x
= = = = =
5
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Vy phng trnh c 3 nghim 3log 2; 0x x= =BI TON 5: S DNG PHNG PHP T N PH- DNG 3I. Phng php:Phng php dng n ph dng 3 s dng 2 n ph cho 2 biu thc m trong phng trnh v kholo bin i phng trnh thnh phng trnh tch.II. VD minh ho:VD1: Gii phng trnh: 2 2 23 2 6 5 2 3 74 4 4 1x x x x x x + + + + +
+ = +Gii: Vit li phng trnh di dng: 2 2 2 23 2 2 6 5 3 2 2 6 54 4 4 .4 1x x x x x x x x + + + + + ++ = +
t
2
2
3 2
2 6 5
4, , 0
4
x x
x x
uu v
v
+
+ +
= >=
Khi phng trnh tng ng vi:
( ) ( )1 1 1 0u v uv u v+ = + =
2
2
3 2 2
22 6 5
1
1 4 1 3 2 0 2
1 12 6 54 1
5
x x
x x
x
u x x x
v xx x
x
+
+ +
= = = + = = = = + + = =
Vy phng trnh c 4 nghim.
VD2: Cho phng trnh:2 25 6 1 6 5.2 2 2.2 (1)x x x xm m + + = +
a) Gii phng trnh vi m=1b) Tm m phng trnh c 4 nghim phn bit.
Gii: Vit li phng trnh di dng:
( )2 22 2 2 2
2 2 2 2
( 5 6) 15 6 1 7 5 5 6 1
5 6 1 5 6 1
.2 2 2 .2 2 2
.2 2 2 .2
x x xx x x x x x x
x x x x x x
m m m m
m m
+ + + +
+ +
+ = + + = +
+ = +
t:
2
2
5 6
1
2, , 02
x x
x
uu v
v
+
= > =
. Khi phng trnh tng ng vi:
( ) ( )2
2
2
5 6
11
31 2 1
1 0 22
2 (*)
x x
x
x
xu
mu v uv m u v m xv m m
m
+
== = + = + = = = = =
Vy vi mi m phng trnh lun c 2 nghim x=3, x=2a) Vi m=1, phng trnh (*) c dng:
21 2 22 1 1 0 1 1x x x x = = = = Vy vi m=1, phng trnh c 4 nghim phn bit: x=3, x=2, x= 1b) (1) c 4 nghim phn bit (*) c 2 nghim phn bit khc 2 v 3.
(*) 2 22 2
0 0
1 log 1 log
m m
x m x m> >
= = . Khi iu kin l:
( )22
2
00 2
1 log 0 1 11 0;2 \ ;1 log 4 8 2568
11 log 9
256
m
m m
mmm
m
mm
>>
6
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Vy vi ( )1 1
0;2 \ ;8 256
m
tho mn iu kin u bi.
BI TON 6: S DNG PHNG PHP T N PH- DNG 4I. Phng php:Phng php dng n ph dng 4 l vic s dng k n ph chuyn phng trnh ban u thnh 1 hphng trnh vi k n ph.Trong h mi th k-1 th phng trnh nhn c t cc mi lin h gia cc i lng tng ng.
Trng hp c bit l vic s dng 1 n ph chuyn phng trnh ban u thnh 1 h phng trnhvi 1 n ph v 1 n x, khi ta thc hin theo cc bc:Bc 1: t iu kin c ngha cho cc biu tng trong phng trnh.
Bc 2: Bin i phng trnh v dng: ( ), 0f x x =
Bc 3: t ( )y x= ta bin i phng trnh thnh h:( )
( ); 0y x
f x y
=
=II. VD minh ho:
VD1: Gii phng trnh:1 1 1
8 2 18
2 1 2 2 2 2 2
x
x x x x + =+ + + +
Gii: Vit li phng trnh di dng: 1 1 1 18 1 18
2 1 2 1 2 2 2x x x x + =
+ + + +
t:1
1
2 1, , 1
2 1
x
x
uu v
v
= + >= +
Nhn xt rng: ( ) ( )1 1 1 1. 2 1 . 2 1 2 2 2x x x xu v u v = + + = + + = +Phng trnh tng ng vi h:
8 1 18 28 18
99;
8
u vu v
u v u vu v uv u v
u v uv
= = + =+ = + + = = = + = + Vi u=v=2, ta c:
1
1
2 1 21
2 1 2
x
xx
+ = =+ =
+ Vi u=9 v9
8v = , ta c:
1
1
2 1 949
2 18
x
xx
+ = =
+ =Vy phng trnh cho c cc nghim x=1 v x=4.
VD2: Gii phng trnh: 22 2 6 6x x + =Gii: t 2xu = , iu kin u>0. Khi phng trnh thnh: 2 6 6u u + =
t 6,v u= + iu kin 26 6v v u = +Khi phng trnh c chuyn thnh h:
( ) ( ) ( )2
2 2
2
6 00
1 06
u v u vu v u v u v u v
u vv u
= + = = + = + + == +
+ Vi u=v ta c:2 36 0 2 3 8
2(1)x
uu u x
u
= = = = =
+ Vi u+v+1=0 ta c:
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2
2
1 21
21 1 21 125 0 2 log
2 21 21(1)
2
x
u
u u x
u
+= + = = =
=
Vy phng trnh c 2 nghim l x=8 v x=2
21 1log .
2
BI 7: S DNG TNH CHT N IU CA HM SI. Phng php:S dng cc tnh cht ca hm s gii phng trnh l dng ton kh quen thuc. Ta c 3 hng pdng:Hng1: Thc hin cc bc sau:
Bc 1: Chuyn phng trnh v dng: f(x)=kBc 2: Xt hm s y=f(x). Dng lp lun khng nh hm s n iu( gi s ng bin)Bc 3: Nhn xt:
+ Vi ( ) ( )0 0x x f x f x k = = = do 0x x= l nghim+ Vi ( ) ( )0x x f x f x k > > = do phng trnh v nghim
+ Vi ( ) ( )0 0x x f x f x k < < = do phng trnh v nghim.Vy 0x x= l nghim duy nht ca phng trnh.Hng 2: Thc hin theo cc bc:
Bc 1: Chuyn phng trnh v dng: f(x)=g(x)Bc 2: Xt hm s y=f(x) v y=g(x). Dng lp lun khng nh hm s y=f(x) l
L ng bin cn hm s y=g(x) l hm hng hoc nghch bin
Xc nh 0x sao cho ( ) ( )0 0f x g x=Bc 3: Vy phng trnh c nghim duy nht 0x x=
Hng 3: Thc hin theo cc bc:Bc 1: Chuyn phng trnh v dng: f(u)=f(v) (3)Bc 2: Xt hm s y=f(x). Dng lp lun khng nh hm s n iu ( gi s
ng bin)Bc 3: Khi : (3) u v = vi , fu v D
II. VD minh ho:VD1: Gii phng trnh: 2log2.3 3xx + = (1)Gii: iu kin x>0. Bin i phng trnh v dng: 2log2.3 3x x= (2)Nhn xt rng:+ V phi ca phng trnh l mt hm nghch bin.+ V tri ca phng trnh l mt hm ng bin.Do vy nu phng trnh c nghim th nghim l duy nht.
Nhn xt rng x=1 l nghim ca phng t rnh (2) v2log2.3 3 1x =
Vy x=1 l nghim duy nht ca phng trnh.
VD2: Gii phng trnh: ( )23 1
23
1log 3 2 2 2
5
x x
x x
+ + + =
(1)
Gii: iu kin:2 13 2 0
2
xx x
x
+
t 2 3 2u x x= + , iu kin 0u suy ra: 2 2 2 23 2 3 1 1x x u x x u + = =
8
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Khi (1) c dng: ( )21
3
1log 2 2
5
u
u
+ + =
Xt hm s: ( ) ( )21
2
3 3
1 1( ) log 2 log 2 .5
5 5
x
f x x x x
= + + = + +
+ Min xc nh [ 0; )D = +
+ o hm: ( )2
1 1 .2 .5 . ln3 0,2 ln 3 5xf x x D
x= + > + . Suy ra hm s tng trn D
Mt khc ( ) ( )31
1 log 1 2 .5 2.7
f = + + =
Do , phng trnh (2) c vit di dng:
( ) ( ) 2 3 51 1 3 2 12
f u f u x x x
= = + = =
Vy phng trnh c hai nghim3 5
2x
=
VD2: Cho phng trnh:22 2 4 22 2 2
5 5 2
x mxx mx
x mx m
+ ++ +
= + +a) Gii phng trnh vi
4
5m =
b) Gii v bin lun phng trnhGii: t 2 2 2t x mx= + + phng trnh c dng: 2 25 5 2 2t t mt t m+ + = + + (1)Xc nh hm s ( ) 5tf t t= ++ Min xc nh D=R+ o hm: 5 . ln 5 1 0,tf x D= + > hm s tng trn D
Vy (1) ( ) ( ) 22 2 2 2 2 0 2 0f t f t m t t m t m x mx m = + = + + = + + = (2)
a) Vi 45
m = ta c: 2 22
8 4 0 5 8 4 0 25 5
5
x
x x x xx
=+ = = =
Vy vi4
5m = phng trnh c 2nghim
22;
5x x= =
b) Xt phng trnh (2) ta c: 2' m m = + Nu 2' 0 0 0 1m m m < < < < . Phng trnh (2) v nghim phng trnh (1) v nghim.+ Nu ' 0 = m=0 hoc m=1.
vi m=0 phng trnh c nghim kp x=0vi m=1 phng trnh c nghim kp x0=-1
+ Nu 1' 0 0mm
> >
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I. Phng php:Vi phng trnh c cha tham s: f(x,m)=g(m). Chng ta thc hin cc bc sau:Bc 1: Lp lun s nghim ca (1) l s giao im ca th hm s (C): y=f(x,m) v ng thng(d): y=g(m).Bc 2: Xt hm s y=f(x,m)+ Tm min xc nh D+ Tnh o hm y ri gii phng trnh y=0
+ Lp bng bin thin ca hm sBc 3: Kt lun:
+ Phng trnh c nghim ( ) ( )min , ( ) max , ( )f x m g m f x m x D + Phng trnh c k nghim phn bit (d) ct (C) ti k im phn bit+ Phng trnh v nghim ( ) ( )d C = III. VD minh ho:
VD1: Cho phng trnh: ( )22 2 2 22 2 23 2 2 2
x xx x x x m + + + + =
a) Gii phng trnh vi m=8b) Gii phng trnh vi m=27c) Tm m phng trnh c nghim
Gii: Vit li phng trnh di dng: 2 22 2 2 2 23 4 2 2x x x x x x m + ++ + + =S nghim ca phng trnh l s giao im ca th hm s:
2 22 2 2 2 23 4 2 2x x x xy x x + += + + + vi ng thng y=m
Xt hm s2 22 2 2 2 23 4 2 2x x x xy x x + += + + + xc nh trn D=R
Gii hn: lim y = +Bng bin thin: v 3>1, 4>1 nn s bin thin ca hm s ph thuc vo s bin thin cca hm s
2 2 2t x x= + ta c:a) Vi m=8 phng trnh c nghim duy nht x=1b) Vi m=27 phng trnh c 2 nghim phn bit x=0 v x=2c) Phng trnh c nghim khi m>8
VD2: Vi gi tr no ca m th phng trnh:2 4 3
4 21 15
x x
m m
+ = +
c 4 nghim phn bit
Gii: V 4 2 1 0m m + > vi mi m do phng trnh tng ng vi:
( )2 4 215
4 3 log 1x x m m + = +
t ( )4 215
log 1m m a + = , khi : 2 4 3x x a + =
Phng trnh ban u c 4 nghim phn bit phng trnh (1) c 4 nghim phn bit
ng thng y=a ct th hm s2
4 3y x x= + ti 4 im phn bit
Xt hm s:2
2
2
4 3 1 34 3
4 3 1 3
x x khix hoacxy x x
x x khi x
+ = + = +
o hm:2 4 1 3
'2 4 1 3
x khix hoacxy
x khi x
< >= + <
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T , ng thng y=a ct th hm s2 4 3y x x= + ti 4 im phn bit
( )4 2 4 215
10 1 0 log 1 1 1 1 0 1
5a m m m m m < < < + < < + < < phng trnh c vit di dng:
2
2
33 1
1
tt m t m
t
++ = + =
+(1)
S nghim ca (1) l s giao im ca th hm s (C): 23
1
ty
t
+=
+vi ng thng (d):y=m
Xt hm s: 23
1
ty
t
+=
+xc nh trn ( )0;D +
+ o hm:
( )2 2
1 3 1' ; ' 0 1 3 0
31 1
ty y t t
t t
= = =
+ ++ Gii hn: ( )lim 1y t= ++ Bng bin thin:
Bin lun:Vi 1m hoc 10m > phng trnh v nghimVi 1 3m< hoc 10m = phng trnh c nghim duy nhtVi3 10m< < phng trnh c 2 nghim phn bit
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CH II:BT PHNG TRNH MBI TON I: S DNG PHNG PHP BIN I TNG NGI. Phng php:Ta s dng cc php bin i tng ng sau:
Dng 1: Vi bt phng trnh:( ) ( ) ( ) ( )
( ) ( )
1
0 1
f x g x
a
f x g xa a
af x g x
>
= <
Ch : Cn c bit lu ti gi tr ca c s a i vi bt phng trnh m.
II. VD minh ho:VD1: Gii cc bt phng trnh:
a) 21
2
12
2
x
x x
b) ( ) ( )3 11 310 3 10 3
x x
x x
+ ++ < +
Gii:a) Bin i tng ng bt phng trnh v dng:
( )
2 22 12
22
1 0
2 01 12 1 21 02 2
2 1
x x x
x
x x
x x x xx
x x x
>
Vy nghim ca bt phng trnh l 2x Ch : trnh sai st khng ng c khi bin i bt phng trnh m vi c s nh hn 1 cc emhc sinh nn la chn cch bin i:
2
2
1 2 1 2 2
2
12 2 2 2 1 2 1 2
2
x x x x
x xx x x x x x x
b) Nhn xt rng: ( ) ( ) ( ) 110 3 10 3 1 10 3 10 3
+ = = +
Khi bt phng trnh c vit di dng:
( ) ( ) ( )
( ) ( )
3 1 3 11 3 1 3
2
10 3 10 3 10 3 1
3 53 1 50 0
1 3 1 3 1 5
x x x x
x x x x
xx x x
x x x x x
+ ++ + ++ + + 0)( )
( )
1
log
0 1log
a
a
a
f x b
af x b
> > > < > hoc c ths dng logarit theo c s a hay b.II. VD minh ho:VD: Gii bt phng trnh:
2
49.2 16.7x x>Gii: Bin i tng ng phng trnh v dng: 4 22 7x x >Ly logarit c s 2 hai v phng trnh ta c:
( )2 4 2 2 2
2 2 2 2 2log 2 log 7 4 2 log 7 ( ) log 7 2 log 7 4 0x x x x f x x x > > = + >
Ta c: ( ) ( ) 222 2 2 2log 7 8log 7 16 log 7 4 4 log 7 = + = = . Suy ra f(x)=0 c nghim:
( ) 12 2
1,22 2 1
2log 7 4 log 7
log 7 22
xx
x x
= = = 2 hoc 2log 7 2x <
BI TON 3: S DNG PHNG PHP T N PH- DNG 1I. Phng php:Mc ch chnh ca phng php ny l chuyn cc bi ton cho v bt phng trnh i s quenbit c bit l cc bt phng trnh bc 2 hoc cc h bt phng trnh.II. VD minh ho:
VD1: Gii bt phng trnh : ( ) ( ) ( )22
2 2 2 2 1 2 1x x x < + Gii: iu kin 2 1 0 0x x .t 2 1xt= , iu kin 0t , khi : 22 1x t= + . Bt phng trnh c dng:
( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( )
2 22 22 2 2 2
2 2 2 2 22 2 2
1 2 1 2 1 1 3 1
1 3 1 0 1 1 3 0
t t t t t t
t t t t t t
+ < + + < +
+ < + +
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( ) ( ) ( )2 31 2 2 0 1 1
2 1 1 2 2 1x x
t t t t
x
< (*)
t 5xu = , iu kin u>2, khi bt phng trnh c dng: 22
3 54
uu
u+ >
(1)
Bnh phng 2 v phng trnh (1) ta c:
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2 2 2 22
2 22 2
4 445 4. 45
4 44 4
u u u uu
u uu u+ + > + >
(2)
t2
2, 0
4
ut t
u= >
. Khi bt phng trnh (2) c dng:
22 4 2
2
2 5
22
4 45 0 5 5 25 100 0
4log 2020 20 5 20(*)
15 log 55 5 5
2
x
x
ut t t u u
ux
u u
u xu
+ > > > + >
> > > > < <
Vy nghim ca bt phng trnh l ( )5 51log 2; log 20;2x +
BI TON 4: S DNG PHNG PHP T N PH- DNG 2I. Phng php:Phng php ny ging nh phng trnh m.II. VD minh ho:
VD1: Gii bt phng trnh: 214 2 4 0x x x+ + Gii: t 2xt= iu kin t>0Khi bt phng trnh c dng:
22 2 4 0xt t + . Ta c:2
' 1 4 0x =
Do :
22' 004 11 4 0
(2) 001 2 1
2
xx
x
xxb
xt ta
= == = = == = = Vy bt phng trnh c nghim duy nht x=0.
VD2: Gii bt phng trnh : ( ) ( )9 2 5 .3 9 2 1 0x xx x + + + Gii: t 3xt= iu kin t>0. khi bt phng trnh tng ng vi:
( ) ( ) ( )2 2 5 9 2 1 0f t t x t x= + + + . Ta c ( ) ( ) ( )2 2' 5 9 2 1 4x x x = + + = .Do f(t)=0 c 2 nghim t=9 hoc t=2x+1
Do bt phng trnh c dng: ( ) ( )9 2 1 0t t x
3 99 0 2
2 1 0 3 2 1 0 1 2
0 19 0 23 92 1 0 0 13 2 1
x
x
x
x
t x
t x x Bemouli x x x
xt x
t x xx
+ +
Vy bt phng trnh c nghim 2x hoc 0 1x
BI TON 5: S DNG PHNG PHP T N PH- DNG 3I. Phng php:S dng 2 n ph cho 2 biu thc m trong bt phng trnh v kho lo bin i bt phng trnhthnh phng trnh tch, khi lu :
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0
0. 0
0
0
A
BA B
A
B
> >> 0 v 0v . Khi bt phng trnh c bin i v dng:
( ) ( ) ( )2 22 2 2 2
2 2 2 2 02 2 1x
u v u v u v u v u v
u v x+ < + + < + >
+
Ta xt phng trnh: 20
2 02 2 1 2 2 1 1
2 12
x x
xx
x xx x
== = + = + = =
Vy bt phng trnh c nghim1 1
; / 0;2 2
x +
VD3:Bt phng trnh : 52 log 2 15 1 5 3 5 2.5 16xx x x+ + + + c nghim la) 1x
b) x>1Gii: Vit li bt phng trnh di dng:
( ) ( )
2 1
2
5 1 5 3 2.5 10.5 16
5 1 5 3 2 5 3 2 5 1
x x x x
x x x x
+ + +
+ +
iu kin: 5 1 0 0x x . t5 1 0
5 3
x
x
u
v
=
= . Bt phng trnh c bin i v dng:
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( ) ( )2 2
2 22 2
2
0 02 2 5 1 5 3
2 2 0
5 3 0 5 31
5 7.5 10 05 1 5 3
x x
x x
x xx x
u v u vu v u v u v
u v u v u v
x
+ + + + = = + +
= + = =
Vy bt phng trnh c nghim x=1.
CC BT PHNG TRNH M C GII BNG NHIU CCHI. T VN :Nh vy thng qua cc bi ton trn, chng ta bit c cc phng php c bn gii btphng trnh m v thng qua cc v d minh ho chng ta cng c th thy ngay mt iu rng, mtbt phng trnh c th c thc hin bng nhiu phng php khc nhau. Trong mc ny s minhho nhng v d c gii bng nhiu phng php khc nhau vi mc ch c bn l:+ Gip cc em hc sinh tip nhn y kin thc ton THPT tr nn linh hot trong vic lachn phng php gii.
+ Gip cc em hc sinh lp 10 v 11 la chn c phng php ph hp vi kin thc ca mnh.II. VD minh ho:VD: Tm m dng bt phng trnh sau c nghim:
( ) ( )2 2 2 22 1 2 1
2 3 2 3 8 4 3x x m m m x x m m m+ + + + + + +
+ + +
Gii: Nhn xt rng: ( ) ( )2 3 . 2 3 1+ =Nn nu t ( )
2 222 3
x x m m m
u+ + +
= + iu kin u>1
Th ( )2 22 1
2 3x x m m m
u
+ + + = . Khi bt phng trnh c dng:
Ta c th la chn 1 trong 2 cch gii sau:Cch 1: S dng phng php t n ph.
t t=x-m, bt phng trnh c dng: ( )2 22 2 1 0t t mt m m+ + + + (2)+ Vi 0t th (2) ( ) ( )2 22 1 2 1 0f t t m t m m = + + + + (3)Vy (2) c nghim (3) c t nht 1 nghim 0t
f(t)=0 c t nht 1 nghim 0t 1 2(0 t t hoc 1 20 )t t
( ) 2 22
2
1 2
11 2 1 0' 022 1 0(0) 0 1
111 0 2
0 12
2 1 0 1(0) 0 12
m
m m mm
m mafmm
msm
m maf m
+ + + +
17
( ) ( )( )
2 2
2
22 2
2 32 3 4 2 3 4 1 0
2 3 2 3 2 3 2 3 2 1(1)x x m m m
u u uu
u x x m m m+ + +
++ + + +
+ + + + + +
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+ Vi 0t th (2) ( )2 2( ) 2 1 2 1 0g t t m t m m = + + + (3)Vy (2) c nghim (3) c t nht 1 nghim 0t
phng trnh g(t)=0 c t nht (1) nghim 1 21 2
00
0
t tt
t t
( ) 2 22
2
1 21 2 1 0' 0
12 1 0(0) 0 12 11 0 1 2
02 12 1 0 1(0) 0 2
mm m m
mm magm
ms m
m m mag
+ + +
Vy bt phng trnh c nghim khi1
02
m<
Cch 2: S dng phng php t n ph
t t x m= , iu kin 0t . Bt phng trnh c dng: 2( ) 2 2 1 0h t t t mx m= + + + (4)Vy bt phng trnh c nghim min ( ) 0( 0)h t t (5)Nhn xt rng h(t) l 1 Parabol c nh t=-1
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CH 3: H PHNG TRNH MBI TON 1: S DNG PHNG PHP T N PHI. Phng php:Phng php c s dng nhiu nht gii cc h m l vic s dng cc n ph. Tu theo dngca h m la chn php t n ph thch hp.Ta thc hin theo cc bc sau:Bc 1: t iu kin cho cc biu thc trong h c ngha
Bc 2: La chn n ph bin i h ban u v cc h i s bit cch gii ( h bc nht 2 n,h i xng loi I, h i xng loi II v h ng cp bc 2)Bc 3: Gii h nhn cBc 4: Kt lun v nghim cho h ban u.II. VD minh ho:
VD1: Gii h phng trnh:2 2 2 2
1
3 2 17
2.3 3.2 8
x y
x y
+ +
+
+ =
+ =(I)
Gii: t3
2
x
y
u
v
=
=iu kin u, v>0. Khi h (I) c bin i v dng:
2
2 21
19 6 1 0 3 19 4 17 338 6 16 3 8 2 2 23
x
y
u u xuu vu
yu v vv
+ = = = =+ = =+ = = = = Vy h c cp nghim (-1;1)
VD2: Cho h phng trnh:1
1
3 2 2
3 2 1
x y
x y
m m
m m
+
+
+ =
+ = +a) Tm m h c nghim duy nht.b) Tm m nguyn nghim duy nht ca h l nghim nguyn.
Gii: t13
2
x
y
u
v
+ =
=iu kin u 3 v v>0. Khi h (I) c bin i v dng:
2
1
mu v m
u mv m
+ = + = +
(II). Ta c:
1
mD = 2
11m
m= ;
2
1um
Dm
=+
21 2 1;
1vm
m m Dm
= = 22
1
mm m
m=
+a) H c nghim duy nht khi:
20 1 01
2 13 3 2 1 2 1
1 1 00
1
u
v
D mm
D mu m m
D m m mD m
vD m
+ = < + < = > +
Vy h c nghim khi 2 1m < .a) Vi m nguyn ta c m=-2 khi h c nghim l:
13 03 3 1 1
2 112 2
x
y
u xx
v yy
+ = == + = = === Vy vi m=-2 h c nghim nguyn (0;1)
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VD3: Cho h phng trnh:2 cot sin
sin cot
9 3
9 81 2
gx y
y gx m
+ =
=a) Gii h phng trnh vim=1
b) Tm m h c cp nghim (x;y) tho mn 02
y
Gii: Bin i h v dng:
2
. 3
u v m
u v
+ =
= Khi u, v l nghim ca phng trnh 2( ) 2 3 0f t t mt = = (1)a) Vi m=1 ta c:
sin0; 02
2cot
1 3 9 32 3 0
3 1 9 1
y
u v
gx
t ut t
t v
> < = = = = = = =
261 ; 2
sin 5 2 6 ; ,2256
cot 0 ; 22 6
2
y k
x l y y k y
k l Zy k
gx x l y y k x l
= + = + = = + = = + = = + = = +
= +Vy vi m=1 h c 2 h cp nghim.
VD4: Gii h phng trnh:
2 2
2
2 2 2
2 2 2
4 2 4 1
2 3.2 16
x x y y
y x y
+
+ +
+ =
=
Gii: Vit li h phng trnh di dng:( )2 2
2
2 1 1 2
2 1
4 4.4 .2 2 1
2 3.4 .2 4
x x y y
y x y
+ = =
(I)
t
2 1
42
x
yuv
= =iu kin 14
u v v>0.
Khi h (I) c bin i v dng:2 2
2
4 1(1)
4 4(2)
u uv v
v uv
+ =
=(II)
gii h (II) ta c th s dng 1 trong 2 cch sau:Cch 1: Kh s hng t do t h ta c: 2 24 13 3 0u uv v + = (3)
t u=tv, khi : ( )2 23
(3) 4 13 3 0 1
4
t
v t tt
= + = =
+ Vi t=3 ta c u=3v do : 2(2) 8 4v = v nghim.
+ Vi1
4t= ta c
14
4u v v u= = do : 2(2) 4 4 1u u = =
2 211 11 04 1
4 222 4
x
y
u xx
v yy
= = == = === Vy h phng trnh c 2 cp nghim (1;2) v (-1;2)Cch 2: Nhn xt rng nu (u;v) l nghim ca h th 0u
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T (2) ta c2 4
3
vu
v
= (4). Thay (4) vo (1) ta c: 4 22 31 16 0v v = (5)
t 2 , 0t v t= > ta c: 2 216
1(5) 2 31 16 0 16 41
4(1)2
tu
t t v vvt
= = = = = == 2 21 11 04 1
222 4
x
y
xx
yy
= == === Vy h phng trnh c 2 cp nghim (1;2) v (-1;2)
VD5: Gii h phng trnh:2 1 2
22
2 3.2 2
2 3 2 2
x x
x
y
y y
+ = =
= Gii: t 2 xu = iu kin 1u . H c dng:
( ) ( ) ( )
( ) ( )
2 22 2 2 2
2 2
2 3 22 3
2 3 2
3 1 0 1
u u yu y u y u y
y y u
u y
u y u y y u
= = =
= + = =
+ Vi u=y, h phng trnh tng ng vi:
2 2 2
2 1 0
1 11
22 3 2 3 2 0 12 222
x
x
x
y yu y u y u y
u yu u u u u x
yy
= = = == = = = = = = + = = = ==
+ Vi y=1-u, h phng trnh tng vi:
( ) 2 221 1
3 1 02 3 1 2
y u y u
u uu u u
= =
+ = = v nghimVy h c 3 cp nghim l (0;1), (1;2) v (-1;2).
VD6: Gii phng trnh:
( ) ( )
( ) ( )
22log 3log
2 2
9 3 2 (1)
1 1 1(2)
xyxy
x y
=
+ + + =Gii: iu kin xy>0
+ Gii (1): t ( )2log 2tt xy xy= = . Khi phng trnh (1) c dng:
( ) 2log 3 2 29 3 2 2 3 3 2.3 3 2.3 3 0t t t t t t = = = (3)t 3 , 0tu u= > , khi phng trnh (3) c dng:
2 1(1)2 3 0 3 3 1 2
3t
uu u t xy
u
= = = = = =
+ Gii (2): ( ) ( )22 2 2 2 1 0 2 2 1 0x y x y x y x y xy + + + + = + + + + =
( ) ( )2 2 3 0x y x y + + + = (4)t v=x+y, khi phng trnh (4) c dng:
21 1
2 3 03 3
v x yv v
v x y
= + = + = = + =
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Vi x+y=1 ta c:1
2
x y
xy
+ = =
Khi x, y l nghim ca phng trnh: 2 2 0X X + = v nghim
Vi x+y=-3, ta c:3
2
x y
xy
+ = =
Khi x, y l nghim ca phng trnh :2 1 1
3 2 0 2 2
X x
X X X y
= = + = = = v
2
1
x
y
= =
Vy h c 2 cp nghim (1;2) v (2;1)
VD7: Gii h phng trnh:
3 1 2 3
2
2 2 3.2 (1)
3 1 1(2)
x y y x
x xy x
+ + + =
+ + = +Gii:
Phng trnh (2) ( )2
1 011 0
0 13 1 03 1 1
3 1 0 1 3
x xxx
x xx x yx xy x
x y y x
= + = + =+ + = + + = =
+ Vi x=0 thay vo (1) ta c: 2 28 82 2 3.2 8 2 12.2 2 log11 11y y y y y y+ = + = = =
+ Vi1
1 3
x
y x
=
thay y=1-3x vo (1) ta c: 3 1 3 12 2 3.2x x+ + = (3)
t 3 12 xt += v 1t nn1
4t
( ) ( )
2 3 1
2 2
3 8(1)1(3) 6 6 1 0 2 3 8
3 8
1
log 3 8 1 2 log 3 83
xt
t t tt t
x y
+ =
+ = + = = += +
= + = +
Vy h phng trnh c 2 nghim:2
0
8log
11
x
y
=
=v
( )( )
2
2
1log 3 8 1
3
2 log 3 8
x
y
= + = +
BI TON 2: S DNG PHNG PHP HM SI. Phng php:Ta thc hin theo cc bc sau:Bc 1: t iu kin cho cc biu thc trong h c ngha.
Bc 2: T h ban u chng ta xc nh c 1 phng trnh h qu theo 1 n hoc c 2 n, giiphng trnh ny bng phng php hm s bitBc 3: Gii h mi nhn cII. VD minh ho:
VD1: Gii h phng trnh:2 2
3 3 (1)
12(2)
x y y x
x xy y
=
+ + =Gii: Xt phng trnh (1) di dng: 3 3x yx y+ = + (3)Xt hm s ( ) 3tf t t= + ng bin trn R.
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Vy phng trnh (3) c vit di dng: ( ) ( )f x f y x y= = . Khi h c dng:
2 2 2
2
2 212 3 12
x y x y x y x y
x x yx xy y x
= = = = = = = = + + = =
Vy h phng trnh c 2 cp nghim (2;2) v (-2;-2)
VD2: Gii h phng trnh:2 2 3
2 2 3
x
y
x y
y x
+ = +
+ = +Gii: Bin i tng ng h v dng:
2 2 32 3 3 2 3 3
3 2 2
x
x y
y
x yx y
x y
+ = + + + = + ++ = +
(1)
Xt hm s ( ) 2 3 3tf t t= + + l hm ng bin trn R.Vy phng trnh (1) c vit di dng: ( ) ( )f x f y x y= = .
Khi h thnh:2 2 3 2 3 (2)x xx y x y
x y x
= =
+ = + = (II)
+ Gii (2): Ta on c x=1 v 12 3 1= . V tri l mt hm ng bin cn v tri l hm s nghchbin do vy x=1 l nghim duy nht ca phng trnh ny. Khi h (II) tr thnh:
11
x yx y
x
= = = =
Vy h cho c nghim x=y=1.
VD3: Gii h phng trnh:( ) ( )
2 2
2 2 2 (1)
2(2)
x y y x xy
x y
= +
+ =Gii: Thay (2) vo (1) ta c:
( ) ( )2 2 3 33 3
2 2 2 2
2 2 (3)
x y x y
x y
y x x y xy y x
x y
= + + =
=
Xt hm s ( ) 32tf t t= + ng bin trn R.
Vy phng trnh (3) c vit di dng: ( ) ( )f x f y x y= = . Khi h c dng:
2 2 2
1
1 12 2 2
x y x y x y x y
x x yx y x
= = = = = = = = + = =
Vy h c 2 cp nghim (1;1) v (-1;-1)
BI TON 3: S DNG PHNG PHP NH GII. Phng php:Nhiu bi ton bng cch nh gi tinh t da trn cc:
+ Tam thc bc hai+Tnh cht hm s m+Bt ng thc+..Ta c th nhanh chng ch ra c nghim ca h hoc bin i h v dng n gin hn.II. VD minh ho:
VD: Gii h phng trnh:
2 2
2
1 1
1
2 3 2 2 3
2 .3 1
x y x y
x y
+ = + =
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Gii: t 2 12x
y
u
v
=
=iu kin u>0 v
1
3v . H c dng:
2(1)
1(2)
u v u v
uv
+ + =
=(I)
Bin i (1) v dng:
( ) ( ) ( ) ( )2 2 2 2 2 2 2 2 2 24 2 2 2 2 4 4u v u v u v u v u v u v uv = + + + = + + + =Khi h tng ng vi:
2
2 2
2 21
2 0 2 1 0 01
1 0 13 11
x
y
u v x xu v u v
y yuv
= = = = = = = = = = =
Vy h c 2 cp nghim (0;1) v (0;-1)
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CH 4: H BT PHNG TRNH MBI TON 1: S DNG PHNG PHP BIN I TNG NGI. Phng php:Da vo cc php ton bin i tng ng cho cc bt ng thc trong h bt phng trnh, ta c
th tm c nghim ca h. Php ton thng c s dng l:A B
A C B DC D
+> + > + >Vic la chn phng php bin i tng ng gii h bt phng trnh m thng c thchin theo cc bc sau:Bc 1: t iu kin cc biu thc ca h c nghaBc 2: Thc hin cc php bin i tng chuyn h v 1 bt phng trnh i s bit cch gii.Bc 3: Kim tra tnh hp l cho nghim tm c, t a ra li kt lun cho h.Vi h bt phng trnh m cha tham s thng c thc hin theo cc bc sau:Bc 1: t iu kin cc biu thc ca h c nghaBc 2: Thc hin cc php bin i tng ng ( phng php th c s dng kh nhiu trongphp bin i tng ng ) nhn c t h 1 bt phng trnh 1 n cha tham s.Bc 3: Gii v bin lun theo tham s bt phng trnh nhn c.Bc 4: Kim tra tnh hp l cho nghim tm c, t a ra kt lun cho h.Ch : i vi h bt phng trnh m 1 n thng c gii tng bt phng trnh ca h, ri kt
hp cc tp nghim tm c a ra kt lun v nghim cho h bt phng trnh.II. VD minh ho:
VD1: Gii h bt phng trnh:
2 22 1 2 2
2
2 9.2 2 (1)
2 5 4 3(2)
x x x x
x x x
+ + + +
< + Gii:Gii (1):
2 2 2 22 22.2 9.2 4.2 0 2.2 9 4.2 0x x x x x x x x+ + = + =
t2
2x xt = iu kin 41
2t . Khi phng trnh c dng:
22
2 2
4
42 9 0 2 9 4 0 2 41(1)
2
12 2 0 (3)
2
x x
t
t t tt t
xx x x x
x
=+ = + = = ==
= = =
Gii (2):
( )
2
22
2
5 512 5 0 2 2
4 3 0 1 3 1451
2 5 0 55 2
1424 3 2 5 255 24 28 0
x xx
x x xxx
xx
x x x xx x
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Bc 3: Gii h nhn c t suy ra nghim x; yBc 4: Kim tra tnh hp l cho nghim tm c, t a ra li kt lun cho h.II. VD minh ho:
VD: Gii h bt phng trnh:( )
2
22 2
3
2 2 2 1
log 2 2 0
x y
x y
=
(I)
Gii: t 22
x
yuv
= =; u, v >
t:2
2
x
y
u
v
=
=, iu kin u, v>0. H c bin i v dng:
( )
( )
222 2
2 2 22
1 (1)2 1
2 1 1 (2)
u v mu v v m
u v u m v u m
+ + + + +
+ + + + + (I)
iu kn cn: Gi s h c nghim (u0;v0) suy ra (v0;u0) cng l nghim ca h. Vy h c nghimduy nht th iu kin cn l u0=v0.
Khi : ( ) 22 20 0 0 01 2 2 1 0u u m u u m+ + + + (1)
Ta cn (1) phi c nghim duy nht1
02
m = =
Vy iu kin cn h c nghim duy nht l m=1/2
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iu kin : Vi1
2m = h c dng:
( )
( )
22
22
11
21
12
u v
v u
+ + + +
(II)
( ) ( )2 22 2 2 22 2
1 1 1 2 2 2 2 1 0
2 2 12 2 02 2 2
u v u v u u v v
u v u v
+ + + + + + + + +
+ + + = =
Nhn xt rng1
2u v= = tho mn h (II) suy ra x=y=-1
Vy h c nghim duy nht khi m=1/2.
BI TON 4: S DNG PHNG PHP NH GII. Phng php:Nhiu bt phng trnh nh gi tinh t da trn:+ Tam thc bc 2+ Cc bt ng thc c bn nh: Csi, Bunhiacpxki+ Tnh cht tr tuyt iTa c th nhanh chng ch ra c nghim ca n.II. VD minh ho:
VD1: Gii h bt phng trnh:2 1 2 2 (1)
2 2 2 2 1(2)
x y y y
x y y y
+
+
+
+ = (I)
Gii: iu kin: ( )2 11 2 0 2 1 0
22 2 1 02 2 0 2 1
yy y
y xx y y x
y
x+
(*)
Gii (1):(*) 2 11 22 1 0
2 1 2 0
xyx
y yx y = + = =
=(3)
Thay (3) vo (2) thy tho mn. Vy h c nghim duy nht x=y=0.
VD2: Gii h phng trnh:( )
( )
232 3 log 5 4
2
3 5 (1)
4 1 3 8(2)
x x y
y y y
+ =
+ + Gii:
Gii (1) ta c: ( ) ( )2
3 32 3 log 54 log 5 15 3 3 5 4 1 3
x xyy y
+ = = + (3)
Gii (2) vi 3y ta c: ( ) ( ) 2 24 1 3 8 3 0 3 0y y y y y y + + + + (4)T (3) v (4) suy ra y=-3, khi h thnh:
21
1; 32 3 03
3; 333
xx yx x
xx yy
y
= = = = = = = = =
Vy h phng trnh c 2 cp nghim (-1;-3) v (3;-3).
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CHNG II:PHNG PHP GII PHNG TRNH-BT PHNG TRNH- H LGA RIT.
CH 1: PHNG TRNH LGARITBI TON 1: S DNG PHNG PHP LGARIT HO V A V CNG C SI. Phng php: chuyn n s khi lgarit ngi ta c th lgarit ho theo cng 1 c s c 2 v ca phng trnh,
bt phng trnh. Chng ta lu cc php bin i c bn sau:Dng 1: Phng trnh:
( )0 1
log ( )a b
af x b
f x a
< = =
Dng 2: Phng trnh: ( ) ( ) ( ) ( )0 1
log log0a a
af x g x
f x g x
< = = >Ch : Vic la chn iu kin f(x)>0 hoc g(x)>0 tu thuc vo phc tp ca f(x) v g(x).II. VD minh ho:
VD1: Gii phng trnh: ( ) ( )29 3 32 log log .log 2 1 1x x x= +
Gii: iu kin:
0
2 1 0 0
2 1 1 0
x
x x
x
> + > + >
. Phng trnh c vit di dng:
( ) ( )
( ) ( )
( )
( ) ( )
2
2
3 3 3 3 3 3
2
3 3 3 3 3 3
3
3 3
0
2
1 12 log log .log 2 1 1 log log .log 2 1 1
2 2
log 2 log .log 2 1 1 log 2 log 2 1 1 log 0
log 0 1
log 2 log 2 1 1 0 2 1 2 2 1 1
11
4 2 1 22 2 1 2
x
x x x x x x
x x x x x x
x x
x x x x x
xx
x xx x
>
= + = +
= + + = = =
+ = = + + + ==
+ = ++ = +
0
2
1 1
44 0
xx x
xx x
>
= = = = Vy phng trnh c nghim x=1 v x=4.VD2: Gii phng trnh: 3 4 5log log logx x x+ =Gii: iu kin x>0. Ta bin i v cng c s 3:
4 4 3
5 5 3
log log 3. log
log log 3. log
x x
x x
=
=khi phng trnh c dng:
( )3 4 3 5 3
3 4 5 3
log log 3. log log 3. log
log 1 log 3 log 3 0 log 0 1
x x x
x x x
+ =
+ = = =Vy phng trnh c nghim x=1.
BI TON 2: S DNG PHNG PHP T N PH- DNG 1I. Phng php:Phng php t n ph dng 1 l vic s dng 1 n ph chuyn phng trnh ban u thnh 1phng trnh vi 1 n ph.
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Ta lu cc php t n ph thng gp sau:
Dng 1: Nu t logat x= vi x>0 th:1
log ; logk ka x
x t at
= = vi 0 1x<
Dng 2: Ta bit rng: log logb bc aa c= do nu t logb xt a= th logb at x= . Tuy nhin trong nhiu biton c cha logb xa , ta thng t n ph dn vi logbt x= .VD minh ho:
VD1: Cho phng trnh: ( ) ( )2 4log 5 1 .log 2.5 2x x
m = (1)a) Gii phng trnh vi m=1b) Xc nh m phng trnh c nghim 1x
Gii: Bin i phng trnh v dng:
( ) ( ) ( ) ( )2 2 2 21 log 5 1 .log 2 5 1 log 5 1 . 1 log 5 1 22
x x x xm m = + = iu kin: 5 1 0 5 1 0x x x > > >t ( )2log 5 1xt= . Khi phng trnh c dng: ( ) ( ) 21 2 2 0t t m f t t t m+ = = + = (2)
a) Vi m=1 ta c:( )
( )
22
22
log 5 1 11 5 1 22 0
2 5 1 2log 5 1 2
x x
xx
tt t
t
= = = + = = = =
5
5
log 35 3
55log5
44
x
x
x
x
= = ==
Vy vi m=1 phng trnh c 2 nghim 5 55
log 3; log4
x x= =
b)Vi ( )2 21 5 1 5 1 4 log 5 1 log 4 2 2x xx t = =
Vy phng trnh (1) c nghim 1x (2) c nghim 2t 1 2
1 2
2 (*)
2
t t
t t
(loi (*))
( ). 2 0 4 2 2 0 3a f m m + .Vy vi 3m tho mn iu kin u bi.
VD2: Gii phng trnh: ( ) ( )2 2 22 3 6log 1 .log 1 log 1x x x x x x + =
Gii: iu kin:
2
2
2
1 0
1 0 1
1 0
x
x x x
x x
>
+ >
Nhn xt rng:
( ) ( ) ( ) ( )
12 2 2 21 1 1 1 1x x x x x x x x
+ = = + Khi phng trnh c vit di dng:
( ) ( ) ( )( ) ( ) ( )
1 12 2 2
2 3 6
2 2 2
2 3 6
log 1 .log 1 log 1
log 1 .log 1 log 1
x x x x x x
x x x x x x
+ + = +
+ + = +
s dng php bin i c s: ( ) ( )2 22 2 6log 1 log 6.log 1x x x x+ = + v ( ) ( )2 23 3 6log 1 log 6.log 1x x x x+ = +
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Khi phng trnh c vit di dng:
( ) ( ) ( )2 2 22 6 3 6 6log 6.log 1 .log 6.log 1 log 1x x x x x x+ + = + (1)t ( )26log 1t x x= + . Khi (1) c dng: ( )2 3
2 3
0log 6. log 6. 1 0
log 6. log 6. 1 0
tt t
t
= = =
+
Vi t=0 ( )
22 2
6 2
1
log 1 0 1 1 11
x x
x x x x xx x
+
+ = + = = + Vi 2 3log 6. log 6. 1 0t =
( ) ( )( )
( )
6
6
6 6
6
2 22 3 6 2 3
log 22 23 6
log 22log 2 log 2
log 22
log 6.log 6.log 1 0 log 6.log 1 1
log 1 log 2 1 3
1 3 13 3
21 3
x x x x
x x x x
x xx
x x
+ = + =
+ = + =
+ = = + =
Vy phng trnh c nghim x=1 v ( )6 6log 2 log 2
13 32x
= +
BI TON 3: S DNG PHNG PHP T N PH- DNG 2I. Phng php:Phng php dng n ph dng 2 l vic s dng 1 nph chuyn phng trnh ban u thnh phngtrnh vi 1 n ph nhng cc h s vn cn cha x.Phng php ny thng c s dng i vi nhng phng trnh khi la chnn ph cho 1 biuthc th cc biu thc cn li khng biu din c trit qua n ph hoc nu biu din c thcng thc biu din li qu phc tp.Khi thng ta c 1 phng trnh bc hai theo n ph ( hoc vn theo n x ) c bit s l 1 s
chnh phng.II. VD minh ho:VD1: Gii phng trnh: ( )2 2 2lg lg .log 4 2log 0x x x x + =Gii: iu kin x>0.
Bin i phng trnh v dng: ( )2 2 2lg 2 lg lg 2 lg 0x x x x + + =t t=lgx, khi phng trnh tng ng vi: ( )2 2 22 log . 2 log 0t x t x + + =
Ta c: ( ) ( )2 22 2 22 log 8log 2 logx x x = + = suy ra phng trnh c nghim
2
lg 22 lg 2 100
lg
lglog lg 0 1lg 2
xt x x
x
xt x x x
== = = == = =
Vy phng trnh c 2 nghim x=100 v x=1
BI TON 4: S DNG PHNG PHP T N PH- DNG 3I. Phng php:
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Phng php dng n ph dng 3 s dng 2 n ph cho 2 biu thc lgarit trong phng trnh v bini phng trnh thnh phng trnh tch.II. VD minh ho:
Gii phng trnh: ( ) ( )2 22 2 2log 1 log .log 2 0x x x x x + = Gii:
iu kin
( ) 2
2
1 0
0 1
0
x x
x x
x x
>
> > >
. Bin i phng trnh v dng:
( ) ( )( ) ( )
22
22 2 2
2 22 2 2
log log .log 2 0
2 log log .log 2 0
x xx x x
x
x x x x x
+ =
+ =
t( )22
2
log
log
u x x
v x
=
=. Khi phng trnh tng ng vi:
( ) ( )
( )2 222
12 2 0 1 2 02
1( )log 1 2 0
24log 2 4
uu v uv u vv
x Lx x x x
xxx
x
=+ = = == = = = == =
Vy phng trnh c 2 nghim x=2 v x=4.
BI TON 5: S DNG PHNG PHP T N PH- DNG 4
I. Phng php:Phng php t n ph dng 4 l vic s dng k n ph chuyn phng trnh ban u thnh 1 hphng trnh vi k n ph.Trong h mi th k-1 phng trnh nhn c t cc mi lin h gia cc i lng tng ngII. VD minh ho:
VD1: Gii phng trnh: ( ) ( )2 22 2log 1 3log 1 2x x x x + + =
Gii: iu kin
2
2
2
1 0
1 0 1
1 0
x
x x x
x x
>
+ >
t( )( )
22
22
log 1
log 1
u x x
v x x
= = +
Nhn xt rng: ( ) ( )2 22 2log 1 log 1u v x x x x+ = + + ( ) ( )2 22 2log 1 . 1 log 1 0x x x x= + = =
Khi phng trnh c chuyn thnh:
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( )( )
22
22
2
2
log 1 10 1
3 2 2 2 1 log 1 1
11 52
4
1 2
x xu v u v u
u v v v x x
x xx
x x
= + = = = + = = = + =
= =
+ =Vy phng trnh c nghim x=5/4.
VD2: Gii phng trnh: ( ) ( )2 22 23 log 4 5 2 5 log 4 5 6x x x x+ + + + = (1)
Gii: iu kin ( )( )
2
2 2 5 22
22
4 5 0
3 log 4 5 0 4 5 2 2 4
5 log 4 5 0
x x
x x x x x
x x
+ > + + +
+ 2 29 2 29(*)x +
t( )( )
22
22
3 log 5
5 log 5
u x x
v x x
= + + = +
iu kin , 0u v . Khi phng trnh c chuyn thnh:
( )
( )( )
( )
( )
22 2 22
22
22
22
22
6 26 22 6 6 2 2
8 5 24 28 06 2 8 14
5
3 log 4 5 2
5 log 4 5 2 log2; 2
14 2 14; 3 log 4 55 5 52
5 log 4 55
u v
u vu v u v v
u v v vv vv
x x
x xv u
v v x x
x x
= = + = = = + = + = + = =
+ + = + == = = = + + =
+ =
( )( )
22
22
2 2
121 1212 225 25
12125
4 5 1
121log 4 5
25
4 5 2 4 3 0
34 5 2 4 5 2 02 2 1
x x
x x
x xx x x x
xx x x x
x
+ = + =
= + = + =
= + = + = =
Vy phng trnh c 4 nghim phn bit.
BI TON 6: S DNG PHNG PHP T N PH- DNG 5I. Phng php:
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Phng php t n ph dng 5 l vic s dng 1 n ph chuyn phng trnh ban u thnh 1 hphng trnh vi 1 n ph v 1 n x.Ta thc hin theo cc bc sau:Bc 1: t iu kin c ngha cho cc biu thc trong phng trnh
Bc 2: Bin i phng trnh v dng: ( ),f x x =0
Bc 3: t ( )y x= , ta bin i phng trnh thnh h:( )
( ); 0
y x
f x y
=
=II. VD minh ho:
VD1: Gii phng trnh: 22 2log log 1 1x x+ + = (1)
Gii: t 2logu x= . Khi phng trnh thnh: 2 1 1u u+ + = (2)
iu kin: 21 0
1 11 0
uu
u
+
t 1v u= + iu kin 0 2v 2 1v u = +Khi phng trnh c chuyn thnh h:
( ) ( ) ( )
2
2 22
1 01 0 1 01
u v u vu v u v u v u v u vv u
= + = = + + + = + == + Khi :
+ Vi v=-u ta c:1 5
2 22
1 51 521 0 log 2
21 5(1)
2
u
u u x x
u
= = = =
+=
+ Vi u-v+1=0 ta c:22
2
1log 00
0 11 log 1
2
xxu
u uu x x
=== + = = = = Vy phng trnh c 3 nghim.
BI TON 7: S DNG TNH CHT N IU CA HM SI. Phng php:S dng tnh cht n iu ca hm s gii phng trnh l dng ton kh quen thuc. Ta c 3hng p dng sau:Hng 1: Thc hin theo cc bc:Bc 1: Chuyn phng trnh v dng: f(x)=k (1)Bc 2: Xt hm s y=f(x). Dng lp lun khng nh hm s n iu (gi s ng bin)Bc 3: Nhn xt:
+ Vi ( ) ( )0 0x x f x f x k = = = do 0x x= l nghim+ Vi ( ) ( )0 0x x f x f x k > > = do phng trnh v nghim+ Vi ( ) ( )0 0x x f x f x k < < = do phng trnh v nghim.Vy x=x0 l nghim duy nht ca phng trnhHng 2: Thc hin theo cc bc:Bc 1: Chuyn phng trnh v dng: f(x)=g(x) (2)Bc 2: Xt hm s y=f(x) v y=g(x). Dng lp lun khng nh hm s y=f(x) l ng bin cn hms y=g(x) l hm hng hoc nghch bin.
Xc nh x0 sao cho f(x0)=g(x0)
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Bc 3: Vy phng trnh c nghim duy nht x=x0Hng 3: Thc hin theo cc bc:Bc 1: Chuyn phng trnh v dng: f(u)=f(v) (3)Bc 2: Xt hm s y=f(x). Dng lp lun khng nh hm s n iu (gi s ng bin)Bc 3: Khi (3) u v = vi , fu v D II. VD minh ho:
VD1: Gii phng trnh:
( )( )22 2log 4 log 8 2x x x + = +
Gii: iu kin
2 4 02
2 0
xx
x
> >
+ >. Vit li phng trnh di dng:
( ) ( ) ( )2
22 2 2 2
4log 4 log 2 3 log 3 log 2 3
2
xx x x x x x
x
+ = = =
+Nhn xt rng:
+ Hm s ( )2log 2y x= l hm ng bin+ Hm s y=3-x l hm nghch bin+ Vy phng trnh nu c nghim th nghim l duy nht+ Nhn xt rng x=3 l nghim ca phng trnh
Vy phng trnh c nghim x=3.VD2: Gii phng trnh: ( ) ( )4 2 225log 2 3 2 log 2 4x x x x =
Gii: iu kin:2
2
2 3 0 1 5
2 4 0 1 5
x x x
x x x
> < > > +
. Vit li phng trnh di dng:
( ) ( )( ) ( )
2 225
2 25 4
log 2 3 log 2 4
log 2 3 log 2 4 (1)
x x x x
x x x x
=
=
t 2 2 4t x x= khi (1) ( )5 4log 1 logt t + = (2)
t 4log 4y
y t t= = phng trnh (2) c chuyn thnh h:4 4 1
4 1 5 15 51 5
y yy
y y
y
t
t
= + = + = + = (3)
Hm s ( ) 4 15 5
y y
f y = +
l hm nghch bin
Ta c:+ Vi y=1, f(1)=1 do y=1 l nghim ca phng trnh (3)+ Vi y>1, f(y)
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Chia c 2 v cho 5 0t ta c:4 3
15 5
t t
+ =
(2)
Nhn xt rng:+ V tri ca phng trnh l mt hm nghch bin+ V phi ca phng trnh l mt hm hng+ Do vy nu phng trnh c nghim th nghim l duy nht
+ Nhn xt rng t=2 l nghim ca phng trnh (2) v
2 2
4 3 15 5
+ = Vi 22 log 2 4t x x= = =Vy x=4 l nghim duy nht ca phng trnh
VD4: Gii phng trnh: ( )23 1
23
1log 3 2 2 2
5
x x
x x
+ + + =
(1)
Gii: iu kin 21
3 2 02
xx x
x
+
t 2 2 2 2 23 2; 0 3 2 3 1 1u x x u x x u x x u= + + = =
Khi (1) c dng: ( )21
3
1log 2 2
5
u
u + + =
(2)
Xt hm s ( ) ( ) ( )2
21
3 3
1 1log 2 log 2 .5
5 5
u
uf u u u
= + + = + +
Min xc nh [ )0;D = +
o hm: ( ) ( )21 1
.2 .5 . ln5 0,2 ln 3 5
uf u u u Du
= + > + .
Suy ra hm s ng bin trn D
Mt khc ( ) ( )3 11 log 1 2 .5 25f = + + =
Khi (2) ( ) ( ) 2 3 51 1 3 2 12
f u f u x x x
= = + = =
Vy phng trnh c 2 nghim3 5
2x
=
BI TON 8: S DNG PHNG PHP NH GII. Phng php:II. VD minh ho:
VD1: Gii phng trnh : ( )3 2log 4 5 1x x + + = (1)Gii:Cch 1: Theo bt ng thc Bunhiacpski ta c:
( ) ( ) ( )3 24 5 1 1 4 5 3 2 log 4 5 1x x x x x x + + + + = + + Vy phng trnh c nghim khi v ch khi:
4 5 1
1 1 2
x xx
+= = l nghim duy nht
Cch 2: Theo bt ng thc Csi ta c:
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( ) ( ) ( ) ( ) ( )( )
2
3 2
4 5 4 5 2 4 5 9 4 5 18
4 5 3 2 log 4 5 1
x x x x x x x x
x x x x
+ + = + + + + + + + + =
+ + + +
Vy phng trnh c nghim khi v ch khi:1
4 52
x x x = + = l nghim duy nht ca phng trnh
CH 2: BT PHNG TRNH LGARITBI TON 1: S DNG PHNG PHP BIN I TNG NGI. Phng php: chuyn n s khi loga ngi ta c th m ho theo cng 1 c s c 2 v bt phng trnh. Chngta lu cc php bin i c bn sau:
Dng 1: Vi bt phng trnh: ( ) ( )log loga af x g x >< < < >> < > > <
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VD2: Gii bt phng trnh: ( )2log 5 8 3 2x x x + >Gii:Cch 1: Bt phng trnh tng ng vi:
2
2 2
2
2 22
11 34 8 3 0
5 8 3 20 11 30 1 5 8 3 0 2 50 5 8 3
4 8 3 0
xx
x x xx x x
xx xx xx x x
x x
> > + > > + > < < <
> < <
(*)
Bin i tng ng bt phng trnh v dng:
( ) ( ) ( ) ( )2 2
2
lg 5 1 lg 10. 5 5 1 10. 5
9 3 3 5
x x x x
x x x
> > > > <
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( )
( )
233
333
2
45 1
5 6 035 5
4 5 2 30 5 1
350 35 5
5 6 0
xx
x xx x
x xx
xx x
x x
+ < + > +
> + + >
t 03 1 1xt x t>= > . Khi bt phng trnh (2) c dng:
( ) ( ) ( ) ( )1 03 2 2
3
0
3
2 0 2 0 1 2 0 2 0
2 1 2 1 8 9
0 1 0 0 11 0
t
x
t t t t t t t t t t t
t x x x
t x xx
+ >
>
> > + > >> > > >
< < <
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( ) ( ) ( ) ( )
2 2 23 3 3 3
22 31 1 1 2 2 22 2 2
222 2 2
1 1 2 22 2
log log log log log log 88 8 8 8
log log log log
x x x xx
x x x x
= = = =
= = =
BI TON 4: S DNG PHNG PHP T N PH- DNG 2I. Phng php:
II. VD minh ho:
Gii bt phng trnh: ( )2 33 2 3 2log log 8 .log log 0x x x x + < (1)Gii: iu kin x>0
Bin i phng trnh tng ng v dng: ( )23 2 3 2log 3 log log 3log 0x x x x + + > < < > > < >>
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( ) ( )
3
2
3
2
2 2 0 1 2 0
log 11 0 3
log 22 0 43 4
1 0 3log 1
2 0 4log 2
uv u v u v
xu x
xv xx
u xx
v xx
< > >
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+ Vi -15
Vy bt phng trnh c nghim: ( )1 3
0; 1; 5;2 2
+
CH 3: H PHNG TRNH LGARITBI TON 1: S DNG PHNG PHP BIN I TNGI. Phng php:Ta thc hin theo cc bc sau:Bc 1: t iu kin cho cc biu thc trong h c nghaBc 2: S dng cc php th nhn c t h 1 phng trnh theo n x hoc y (i khi c th ltheo c 2 n x, y)Bc 3: Gii phng trnh nhn c bng cc phng php bit i vi phng trnh cha cnthcBc 4: Kt lun v nghim cho h phng trnh.II. VD minh ho:
VD1: Gii h phng trnh: ( )3
3 41 3 (1)
log 1(2)
y xxx
y x
+ =
+ =
Gii: iu kin:
1 0
4 0 0 4
0
x
x x
x
+ < >
T phng trnh (2) ta c: 33
1 log3 log
3 31 log 3 3
3xy
xy x
x
= = = = (3)
Th (3) vo (1) ta c:
( )
( ) 2 2
3 3 41 1 1 1 4 1 4 1
2 0 24 2 3 03 04 2
xx x x x x
x x
x xx x x y
x xx x
+ = + = + = +
= = = = =
Vy h phng trnh c 1 cp nghim (3;0).
VD2: Gii h phng trnh:( ) ( )
2 2
2 3
4 2
log 2 log 2 1
x y
x y x y
=
+ =
Gii: iu kin:2 0
2 0
x y
x y
+ > >
(*)
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T phng trnh th nht ca h ly lgarit c s 2 hai v ta c:
( ) ( ) ( )( ) ( )
2 22 2 2 2
2 2
log 4 log 2 log 2 log 2 1
log 2 1 log 2
x y x y x y
x y x y
= + + =
+ = Th vo phng trnh th hai ta c:
( ) ( ) ( ) ( )
( )
2 3 2 3 2
2
1 log 2 log 2.log 2 1 1 log 2 log 2 0
log 2 0 2 1
x y x y x y
x y x y
= + =
= =
Vy ta c h mi:2 2
32 24 2 42 1 12 1
2
xx yx y
x yx yy
=+ = = = = =
tho mn iu kin (*)
Vy h phng trnh c 1 nghim.
BI TON 2: S DNG PHNG PHP T N PHI. Phng php:Phng php c s dng nhiu nht gii cc h lgarit l vic s dng cc n ph. Tu theodng ca h m la chn php t n ph thch hp.Ta thc hin theo cc bc sau:Bc 1: t iu kin cho cc biu thc ca h c ngha.Bc 2: La chn n ph bin i h ban u v cc h i s bit cch gii (h i xng loi I,loi II v h ng cp bc hai)Bc 3: Gii h nhn cBc 4: Kt lun v nghim cho h ban u.II. VD minh ho:
Gii h phng trnh:
( ) ( )3 3
4 32
log 1 log
x y
y x
x y x y
+ =
= +
Gii: iu kin:
0
0
; 0
x y
x y
x y
> + >
Bin i h phng trnh v dng:
( )2 2 2 23
2 5 2 5(1)
log 1 3(2)
x y x y
y x y x
x y x y
+ = + = = =
Gii (1): t1x y
t
y x t
= = . Khi (1) c dng:
2
221
2 5 2 5 2 0 12
2
tx y
t t ty xt t
= = + = + = ==
+ Vi x=2y2 2 1 2(2) 4 3
1 2(1)
y xy y
y x
= = = = =
+ Vi y=2x 2 2(2) 4 3x y = v nghimVy h phng trnh c 1 cp nghim (2;1)
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BI TON 3: S DNG PHNG PHP HM SI. Phng phpTa thc hin theo cc bc sau:Bc 1: t iu kin cho 2 biu thc ca h c nghaBc 2: T h ban u chng ta xc nh c 1 phng trnh h qu theo 1 n hoc theo c 2 n, giiphng trnh ny bng phng php hm s bit.Bc 3: Gii h mi nhn c.
II. VD minh ho:
Gii h phng trnh:2 3
2 3
log 3 1 log
log 3 1 log
x y
y x
+ = +
+ = +Gii: iu kin x; y>0. Bin i tng ng h v dng:
( ) ( )( ) ( )
( ) ( )( ) ( )
2 3 2 3
2 3 3 2
log 3 2 1 log log 3 2 1 log
log 3 2 1 log 2 1 log log 3
x y x y
y x x y
+ = + + = + + = + + = +
(I)
( ) ( )2 3 2 3log 3 2log log 3 2logx x y y + + = + + (1)Xt hm s: ( ) ( )2 3log 3 2 logf t t t = + +
Min xc nh ( )0;D = + .o hm ( ) ( )
1 20,
3 ln 2 .ln 3f t t D
t t= + >
+ hm s lun ng bin.
Vy phng trnh (1) c vit di dng: ( ) ( )f x f y x y= =
Khi h (I) tr thmh: ( ) ( )2 3log 3 2 1 log (2)x y
x x
= + = +
(II)
+ Gii (2): ( )2 2
3 3 3 22 1 log log log 2.log3 2 3 4.2 3 4.2x x xx x x+ + = + = + =
( ) 3 3 3 3log 2 log 4 1 log 4 log 423 4. 3 4. 3. 4x x x x x x + = + = + = (3)Xt hm s ( )
3 31 log 4 log 43.g x x x = +Min xc nh ( )0;D = +o hm: ( ) ( ) 3 3log 4 1 log 43 3' 1 log 4 . 3log 4. 0g x x x x D = < hm s lun nghch binVy phng trnh (3) nu c nghim th nghim l duy nhtNhn xt rng nu x=1 l nghim ca phng trnh bi khi :
3 31 log 4 1 log 41 3.1 4 4 4 + = = ng
Khi h (II) tr thnh: 11
x yx y
x
= = = =
Vy h cho c nghim duy nht (1;1)
BI TON 4: S DNG PHNG PHP NH GII. Phng php:II. VD minh ho:
VD1: Gii h phng trnh:( ) ( )2 2
2 2
log log 1 (1)
1(2)
x ye e y x xy
x y
= +
+ Gii: iu kin x; y>0*) Gii (1) ta c nhn xt sau:
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- Nu 2 2log logx y x y> > , khi :( )
( )
1
1
0
0
VT
VP
> + >
+ > + > < + + T phng trnh th nht ca h vi vic s dng n ph t=x+y>0, ta c: 2log 1t t=
t 2log 2uu t t= = khi phng trnh c dng:
2
2
log 00 12 1
1 log 1 2Bernoulliu
tu x yu
u t x y
== + = = + = = + =
+ Vi x+y=1 h c dng: ( )3
1 1 1 0; 1
log 1 0 1 1 0 1; 0
x y x y x y x y
xy xy xy x y
+ = + = + = = = + = + = = = =
+ Vi x+y=2 h c dng: ( )42 2 2
log 1 1 1 4 3x y x y x y
xy xy xy+ = + = + = + = + = =
Khi x; y l nghim ca phng trnh: 2 2 3 0t t + = v nghimVy h c 2 cp nghim (0;1) v (1;0)
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PHNG TRNH M1) 6224 241 +=+ +++ xxx
2) 0273.43 5284 =+ ++ xx
3) 26.52.93.4x
xx =4) xxx 6242.33.8 +=+
5) ( ) 77.0.61007 2
+=x
x
x
6) 13250125 +=+ xxx
7) 623.233.4 212 ++=++ + xxxx xxx
8) 5008.51
=x
x
x
9) 7503333 4321 =++ + xxxx
10) 3421 5353.7 ++++ = xxxx
11) 09.66.134.6 =+ xxx
12) 1284 = xx
13) 1105.35 1212 = + xx
14)xxx
6.59.24.3 =+15) 0273.43 582 =+ ++ xx
16) 3421 5353.7 ++++ = xxxx
17) 04.66.139.61
.611
=++
xxx
18) ( ) ( ) ( )3210101
32321212 22
=++
+ xxxx
19) 02525 21 =++ + xxxx
20) 5332 242 + = xxx
21) 1223
2
1
3229 ++= x
xxx
22) ( ) 329log2 =+xx
23) ( ) ( )( ) ( )3243234732 +=+++ xx
24) xxx 9.21525 =+25) 22 2.10164 =+ xx
26) 022.92 221222
=+ +++ xxxx
27) ( ) 12
12
2
12.62
13
3 =+ xxxx
28) xx
231 2 =+29) 1282
=
x
30) 0624 =+ xx
31) 055.625 31 =+ +xX
32) 073.59 =++ xx
33) 0543.259 = xx
34) 3033 22 =+ + xx
35)( )
093.82312
=++ xx
36) xxxx 3223 7.955.97 +=+37) 033.369 31
22
=+ xx
38) 0639 1122
= ++ xx
39) 123
694 ++
=+ xxx
40) xxxx 3.25.235 22 ++=41) 211
2222
2332 + = xxxx
42) xxx = 21 105
15.2
43) ( ) ( )3
2531653+
=++
xxx
44) xxx 36.281.216.3 =+45) ( ) 2
log
122222
2
xx
xxlo
+=
++
46) 8444242 22 +=+ xxxxx
47) 3loglog29log 222 3. xxx x =
48) 68.3 2 =+xx
x
49) 052.2 82 log3log =+ xx xx50) 5log3log 22 xxx =+
51) ( ) ( ) ( ) 324log 242 2 = xx x
52) xxx 100lglg10lg 3.264 =
53) 626
1
2
12
3
13 +=
+
x
xx
x
x
x
54) 093.613.73.5 1112 =++ + xxxx
55) 20515.33.12 1 =+ +xxx
56)2
222 4log6log2log 3.24xx
x =57) 2653 +=+ xxx
58) ( ) 21 1222
= xxxx
PHNG TRNH LGARIT
1) ( ) ( ) 8log21log3log 444 =+ xx 42) ( xxx 4846 loglog.2 =+
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2) ( 2652log 25 =+ xxx3)
( ) ( ) ( )12lg2021lg110lg5lg =++ xxx
4)
+
+=
8
1lg
2
1
2
1lg
2
1lg
2
1lg xxxx
5) 4lglg3lg 22 = xxx
6)02log3log
31
31 =+
xx
7) ( ) 88
log4log2
2
2
2
1 =+x
x
8) ( ) ( ) 222log64log 255 = xx
9)3
22
4
2
log3log2log4 xxx xxx =+
10) 1log2log2
33 = xx
11) 0log14log40log3
1642
2
=+ xxx xxx
12)
2 3 5
2 3 2 5 3 5
log . log . log
log .log log .log log .log
x x x
x x x x x x= + +
13) ( ) xx
xx 2
3
323 log2
1
3loglog.3log +=
14) ( ) ( 02lglglglg 3 =+ xx15) ( ) ( ) 212log1log 53 =+++ xx
16) ( ) ( ) 01106log3log 222 =+ xx17)
(xxx 4
46 loglog.2 =+
18) 1log 22 = xxx19) 12log.4log 22
2 =xxx20) ( ) 05,4lg1log =+xx
21) 33
log3
log 22 =
+
+
x
x
x
x
22) ( ) ( )2lg46lg 2 ++=+ xxxx23) ( ( ) 01106log3log 222 =+ xx24) 633log33log.log 33 =+xx
25)( ) ( )32log22log 232
2
322= ++ xxxx
26) 112log.loglog.2 332
9 += xxx
27) 013loglog.3 33 = xx
28)
xxxx x 24
2
44
2 log2log2log2log =++
29) ( ) 4lg2lg2
110lg 2 =++ xx
43) ( ) ( ) 61log1log 232
22
32=++++ + xxxx
44) ( ) 34log2log 22 =+ xx
45) ( ) 33logloglog4
3
3
3
13=++ xxx
46) 06
74log2log =+ xx
47) 225log.3logloglog 9535 =+ xx
48) ( ) ( )1log22log
113log 2
3
2 ++=++
xx
x
49) ( ) ( )32log44log1
2
12 =++xx x
50) xxxx 7272 log.log2log2log +=+51)
1log1log.1log 2202
52
4 =+ xxxxxx
52) ( ) 43.59log 2 =+ xx
53) [ ] 1323.49log1
+=+
xxx
x
54) ( 169loglog 3 =xx55) ( ) ( ) 1122log42log 22 +=+ xx x56) ( ) 16log1log 12 +=+ xx
57) ( ) ( ) 2loglog 12222 22 xx xx +=++
58) ( ) ( )1log1log2
1
2
2 = xx
59) ( )( ) 1logloglog 232 =x60)
61) ( ) ( ) ( ) ( ) 01lg.1241lg1 22222 =+++ xxxx62) ( ) ( ) ( ) 0621log51log 3
23 =++++ xxxx
63)
( ( ( 1log1log.1log 262322 =+ xxxxxx64) 5logloglog
3
8
16
14 =++ xxx
65) ( ) ( ) 155log.15log 1255 = +xx66) 225log.3logloglog 9535 =+ xx
67) ( ) ( ) 0226log8log 39 =+++ xx
68)4log.27log.
9
2
+=xxx
x
69) ( ) 944log2log 232
3 =++++ xxx
70)( ) ( ) 02144log156log 231221 =++ xxxx xx
71) ( ) ( ) ( ) ( ) 01lg1241lg1 22222 =+++ xxxx72)
( ) ( ) ( )2log22log5log1log25
15
5
1
2
5 +=++ xxx
73) ( ) ( ) ( ) ( ) 0161log141log2 323 =+++++ xxxx
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30) ( ) ( ) xxx x 2221log2 log1log233
2 +=+
31) ( ) ( ) ( ) ( ) 162log242log3 323 =+++++ xxxx
32) ( ) ( )2
1213log 23 =++ xxx
33)154
22
2
22
3log81log4log
36log =+ xx
34) ( ) 212log2
1 =++ xx35)36) ( ) 062log1log 2
2
2 =++ xxxx37) 33loglog.4 9 =+ xx
38) ( 13log6log 22 = xx39)
3log3127log23log 22
2
2
2 +=+++++ xxxx
40) ( ( ) 0log211 22 =++ xxxx
41) ( ) ( ) 155log.15log1
255 = +xx
74) ( ) ( ) ( )3
4
1
3
4
1
2
4
1 6log4log32log2
3 ++=+ xxx
75) xx
xx
2
3
323 log2
1
3loglog.
3log +=
76)( ) ( ) 421236log4129log 252273 =+++++ ++ xxxx xx
77) ( ) xxxxxxxx 2325log325log. 226
12
62 +=
78) 3logloglog.log 23
332 += xxxx
79)( ) ( )
( )
2
2 1
2
1 2
3 .log 1 2 log 2
3 .log 2 2log 1
x
x
x x
x x
+
= +
80) ( ) ( ) xxx
xxx 2772
2 log3log22
3loglog
++=++
BT PHNG TRNH M1) 922 7 + xx
2) 123
13
3
11
12
=
+
+
xx
3) 4loglog .3416 aa xx +4) ( ) ( ) xxxxxx ++++ +++ xxxx
6) ( ) 13.43 2242
+ xx x
7) 8log.2164 41 +
x
xx
9) 022
1212
32
12 +
++
x
x
10) xxx111
9.46.54.9
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1) 21
18log
2
2 ++
x
xx
2) ( ) ( ) 224log12log 32 +++ xx
3) ( ) 123log2
2
1 + xx
4) ( ) ( )243log1243log 2329 ++>+++ xxxx
5)3
16
5log 3 x
x
x
6)2
1
1
12log 4 ++ xxxx
15) 11
32log 3
x
x
52)2
1122log 2
12 xx
54)( ) ( )
043
1log1log2
3
3
2
2 >
++xx
xx
55) ( ) 22lglg
23lg 2 >+
+x
xx
56) 216_185log2
3>+ xx
x
57) 316log64log 22 + xx
58) 0loglog2
4
1
2
2
1 ++ 2log1244log22
1
2
2
61) ( ) 2log2log 12 ++ xxx62)
( ) ( )1log.112
1log1log.2
5
15225
xx
x
63) ( ) ( )232log1232log 2224 ++>+++ xxxx
64) xx
xx 2
2
122
32
2
1
4
2 log432
log98
loglog + xxx
66) ( ) ( )73log219log1
2
11
2
1 +>+ xx
a. ( ) + 28log x 4x 3 1b. 2
1 4
3
log log x 5 0
48
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27) ( 12log 2 >+ xxx28)
( ) ( )xxx
xx xlog22421412
1272 22 +
++
29) ( ) 2385log 2 >+ xxx30) =31) ( ) 4316 13log.13log
4
14
x
x
32) ( 015log 3,0 >++ xx
33) ( )3log2
12log65log
3
1
3
12
3 +>++ xxxx
34)( ) ( )
0352
114log114log
2
32
11
22
5 >
++
xx
xxx
35) ( )2
3
2
9
4
1loglog
xx
36)2
1
2
54log 2
x
xx
37) ( ) ( )32
1
2
1 21log1log2
1xx >
38) 1log2
1log
2
32
34 > xx
39)( )
014log
5
2
x
x
40) ( )22log1log2
2
2
+
42)( )
082
1log
2
2
1
44) ( 164loglog 2 xx45) xxxx 5353 log.logloglog 5 xlog 3x 4.log 5 1
m. +
+
2
3 2
x 4x 3log 0
x x 5
n.+ >1 3
2
log x log x 1
o. ( ) + 23x xlog 3 x 1
q.+
+ 223x
x 1
5log x x 1 02
r. + > +
x 6 23
x 1log log 0
x 2
s. + 22 2log x log x 0
t. > x x 216
1log 2.log 2
log x 6
u. + 23 3 3log x 4log x 9 2log x 3
v. ( )+ < 2 41 2 162
log x 4log x 2 4 log x
49
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H PHNG TRNH M LGA
I. H phng trnh m.
1)
=
=
+
13
35
4
yx
yx
xy
xy
2)
( ) ( )
+==
+
yxyx
x
y
y
x
33 l o g1l o g3 24
3)
=+
=
423
99.3
1 21
y
x
x
yx
y
x
y
4)( )
=
=
y
y
y
x
x
y
y
x
12
3
5
2
3.33
2.225)
=+
=
2l gl g
1
22yx
x y
6)
=
=
723
723
2
2
y
x
yx
7)
=
+=2
l o g.2
l o gl o g
43
xx y
yy
yx
xy
8)
=
=++ 132
63.22.3
11 yx
yx
9)
=
=+
33
3
3.55
5
yx
yx
yx
yx
10)
=+
+
=+
y
yy
x
xx
x
22
24
452
1
23
11)( )
( )
=+
=+
068
13.4
4
4
4
yx
xy
yx
yx
12)
( ) ( )
=
=3l g4l g
l gl g
34
43
yx
yx
13)
( )
==
1l o g.
3l o g
4
2
5l o g
xyy
x
y
y
xxy
14)
+=++
=+ ++
113
2.322
2
3213
xx yx
xyyx
15)
( ) ( ) ( )
=++=++
++3
81l o g2l o g
1 42
21 xy
yx yx
yx
16)( ) ( )( ) ( )
=+++
=++
421223
421223
xy
yx
17)+
=
=
x y
3x 2y 3
4 128
5 118)
+
= =
2
x y
(x y) 1
5 125
4 1
19) =
=
2x y
x y
3 2 77
3 2 720)
+ =
+ =
x y2 2 12
x y 5
50
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II.H phng trnh lgarit.
1)
( ) ( )
=+
+=
1 6
2l o gl o g
33
22
yx
x yxyyx
2)
( ) ( )
=
=3l g4l g
l gl g
34
43
yx
yx
3)
( )
=+
+=+
1l o g
3l o g2l o gl o g
7
222
yx
yx
4)
=
=+
8
5l o gl o g2
x y
yxxy
5)
=
=+
1l o gl o g
4
44
l o gl o g 88
yx
yxxy
6)
=
=+
1l o gl o g
4
44
l o gl o g 88
yx
yxxy
7)( )
=+
=+
1l o g
433.11
3
yx
x
xx
y
8)
( ) ( )
( ) ( )
=
=
xx
yx
4224
2442
l o gl o gl o gl o g
l o gl o gl o gl o g
15)( )
( )
=+
+=
0l g.l gl g
l gl gl g
2
222
yxyx
x yyx
16)( ) ( )
( ) ( )
=++
=++++
+
+
14l o g5l o g
612l o g22l o g.2
21
2
21
xy
xxyxx y
yx
yx
17)
( ) ( ) ( )
( ) ( )
=+++
+=++
1l o g4224l o g1l o g
3l o g12l o gl o g
4
2
44
44
22
4
y
xxyyx y
yxxyx
18)
( ) ( )
=+
+=
133
24
22
2l o gl o g 33
yxyx
x yx y
19)
( ) ( )
+==
+
yxyx
y
x
x
y
33 l o g1l o g3 24
20)
( )
=+
=+
yyy
yx
x8 13.1 22
3l o g
2
3
21)
=
=
2l o g
4l o g
2
1
2
y
x
x y
51
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9)
( ) ( ) ( )
=++
=++
++3
81l o g2l o g
1 42
21 xy
yx yx
yx
10) ( )( )
=+=+
223l o g223l o g
xy
yx
y
x
11)
( ) ( )
( ) ( )
=++
=+++
453l o g.53l o g
453l o g53l o g
xyyx
xyyx
yx
yx
12)
=+
=
02
0l o gl o g21
23
32
3
yyx
yx
13)
=
=+
1l o gl o g
22
33
l o gl o g 33
xy
yxxy
14)
=
=
9l o gl o g.5
8l o gl o g.5
4
3
2
242
yx
yx
22)
( )
=++
=++
01422
22
32
222
12
2
xyxxyx
x yyx
x
23.+ =
+ =
2 2
lgx lgy 1
x y 29
24.+ = +
+ =
3 3 3log x log y 1 log 2
x y 5
25.( )
( ) ( )
+ = +
+ =
2 2lg x y 1 3lg2
lg x y lg x y lg3
26. =
+ =
4 2
2 2
log x log y 0
x 5y 4 0
27.( ) ( )
+ = + = +
x yy x
3 3
4 32
log x y 1 log x y28.
= = + y
2x y
2log x
log xy log x
y 4y 3
1) (A07) 3 13
2 log (4 3) log (2 3) 2x x + + ( 3 34
x< )
2) (D305) 22 43
log log ( 4 4 )2
x
x x
x
++ + + >
(x>2 4x < )
3) (D206) 2 4 2 12(log 1) log log 04x x+ + = ( x=2 x= )
4) (B203) 0,5 0,25 2log 2 log ( 1) log 6x x+ +
(x 3)
5) 2 4 12
log 2 log 5 log 8 0x x- + + + = 3 176;3;2
x - -
6)2
2 4 1
2
log ( 2) log ( 5) log 8 0x x+ + - + = 3 176;2
x x = =
7) 84 221 1
log ( 3) log ( 1) log (4 )2 4
x x x+ + = (x = 3 x= 3+ 12 )
52
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8)2
9 1 3
3
log ( 3) log 2 log 2 1x x+ - - - < ( 4; 3) ( 3; 1) (0;2) (2;3)- - - -
9) 33log 1 log 12 .5 400x x+ + < ( -10 < x < 8 )
10) (B104)12 4 16
42
x x
x
+ >
(x 4)
11) (A104)2
2
4
log [log ( 2 )] 0x x x
+ < (x >1 x< - 4)
12) (B204) 3log log 3xx > ( x>3 1/3
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38)(A1-08) 1 23
2 3log (log ) 01
+ +
x
xx < 1
39)(A1-08) sin( )4 tanp-
=x
e xx= /4 + k
40)(A2-08)3
1 63 log (9 )log
+ = -x
x
x xx = 2
41)(B1-08) 122
2log (2 2) log (9 1) 1+ + - =x x x= 1; x = 32
42)(B2-08) 2 1 2 13 2 5.6 0+ +- - x x x 23
1log2
x
43)(D1-08)2 22 4 2 2 12 16.2 2 0- - - -- - x x x x 1 3 1- +x
44)(D1-07)2 2
1 22
1 1log 2 3 1 log ( 1)2 2
- + + - x x x x x
48)(A2-07) 4 22 11 1
log ( 1) log 2log 4 2+- + = + +xx x
x =52
49)(B1-07) 23 3log ( 1) log (2 1) 2- + - =x x x=2
50)(B2-07) 3 93
4(2 log )log 3 11 log
+ - =-
xx
xx= 1
3; x= 81
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