Chun đ 1 PHƯƠNG TRNH VƠ TY
* Dạng 1 :
A 0 (hoặc B 0 )
A B
A B
≥ ≥
= ⇔
=
T
n n
A B=
* Dạng 2 :
2
B 0
A B
A B
≥
= ⇔
=
T
n
A B=
Dng 3
A B=
T
n n
A B
+ +
=
Dng 4
A B A B= ⇔ =
n
A B
+
=
+ − = − − = − + −
+ − − = − + + =
+ + − − + =
2 2 2
3 3
1. :
) 4 2 2 (1) ) 4 2 8 12 6 (4)
) 3 1 4 1 (2) ) 12 14 2 (5)
) 11 11 4 (3)
Ví dụ Giảicác phươngtrình
a x x x d x x x x
b x x e x x
c x x x x
( )
− ≥
≥
≥
⇔ ⇔ ⇔ ⇔ =
= ∨ =
− =
+ − = −
=
2
2
2
) :
2 0
2
2
(1) 3
0 3
3 0
4 2 2
: 3
a Tacó
x
x
x
x
x x
x x
x x x
Vậy x
( )
≥ −
⇔ + = + + ⇔ + = + + + + ⇔ + = −
− ≥
≥
≥
⇔ ⇔ ⇔ ⇔ =
= ∨ =
− =
+ = −
=
2
2
1
) :
3
(2) 3 1 1 4 3 1 1 4 2 4 4 2
2 0
2
2
5
0 5
5 0
4 2
: 5
b Tacóđiều kiện x
x x x x x x x
x
x
x
x
x x
x x
x x
Vậy x
( )
± + ≥
⇔ + + + − + + − − = ⇔ − − = −
− ≥
⇔ ⇔ =
− − = −
=
2 2
2
) : 11 0 ( )
(3) 11 11 2 11 16 11 8
8 0
5 ( ( ))
2 11 8
: 5
c Tacóđiều kiện x x a
x x x x x x x x x
x
x thỏa a
x x x
Vậy x
= − + ≥ = − +
− + = − + =
−
⇔ = − ⇔ − = ⇔ = ∨ = ⇔ ⇔ =
− + =
− + =
=
2 2 2
2 2
2
2
2
2
) 2 8 12 0, 2 8 12
2 8 12 0 2 8 12 0
12
(4) 6 2 0 0 2 2
2
2 8 8 0
2 8 12 2
: 2
d Đặt t x x ta có t t x x
x x x x
t
t t t t t x
x x
x x
Vậy x
( )
+ = + + +
⇔ − + + + − + − + + =
− + = −
− + + =
⇔ − + = − ⇔ + − = ⇔ = − ∨ =
= − ∨ =
= − = +
3 3 3
3 3 3 3
3 3
3 3
3 3
) 1: :( ) 3 ( )
(5) 12 14 3 12 . 14 12 14 8
12 . 14 .2 6 ( )
12 14 2 ( )
( ) 12 14 27 2 2 195 0 15 13 ( ( ))
: 15 13
2: 12 ; 14 .
e C Ta có a b a b ab a b
x x x x x x
x x a
x x b
a x x x x x x thỏa b
Vậy x x
C Đặt u x v x Ta
+ = + =
+ = = − =
⇔ ⇔ ⇔ ∨
= − = = −
+ = + − + =
− = − ⇔ =
− = ⇔ =
3 3 3
3
3
:
2 2
2 1 3
3 3 1
26 ( ) 3 ( ) 26
* 12 1 13
* 12 3 15
có
u v u v
u v u u
uv v v
u v u v uv u v
x x
x x
!"#$%
* Phương pháp 1 : Biến đổi về dạng cơ bản
Ví dụ : Giải phương trình sau :
1)
& −=− xx
(x=6) 4)
'(
=−++− xxx
1
(x )
2
= −
2)
)&
−=−+− xxx
(
)
&
=x
) 5)
+=+− xxx
(
*
) ±−
=x
3)
+ =−− xx
(
)
=
x
) 6)
&
&
&
xx
=−
(
±=x
)
* Phương pháp 2 : Đặt điều kiện (nếu có) và nâng luỹ thừa để khử căn thức
1)
&( ++−=+ xxx
(
11
x 0 x )
3
= ∨ =
4)
') =−−−−− xxx
(x=2)
2)
+ =+−− xx
(
(=x
) 5)
, +=−+ xxx
(
=x
)
3)
+=++ xxx
(
+−
=x
) 6)
& +−=+ xx
(
'=x
)
* Phương pháp 3 : Đặt ẩn phụ chuyển về phương trình hoặc hệ pt đại số
1)
xxxx **-)-
+=−+
(x 1 x 4)= ∨ = −
5)
)*&*--& =−++−++ xxxx
(x 0 x 3)= ∨ =
2)
'
=+−+− xxx
(x 1 x 2 2)= ∨ = −
6)
−−=− xx
(x 1 x 2 x 10)= ∨ = ∨ =
3)
&*)*--) =−++−++ xxxx
(
) ±
=x
) 7)
.&&
−+−=−++ xxxx
(x=5)
4)
.
=+−++− xxxx
(x=1; x=2) 8)
)(&
+−+−=−+− xxxxx
(x=2)
Luyn tâp:-!"/01234567898:;#"<;7=>?@A0->?B4
0CD*7*
1EF/G>HC5
*
) - &*x x x− = − =
!*
( - *x x x x− + + = =
*
& + & - I *x x x x x+ + = + = =
*
& - ,*
+
x
x x
x
−
= − =
−
*
& - *x x x x− − = − = −
J*
. ) - *x x x x− = − − − − = −
*
&
& + 'I
x x x x x
− = − − = =
÷
*
&
)
&
x x x x
− − = − =
÷
2EF/G>HC
*
& & - &*x x x− + + = =
!*
I I
x x x x x x
− + − = − = = =
÷
3EF?G>HC
*
( ) ( )
& ) . - +I *x x x x x x+ + − + + = = − =
!*
, ) + I
x x x x x x
− + + − + = = − =
÷
*
+ ( - I *x x x x x x x x+ + + + + = + + = − =
*
, + + - *x x x x x+ − + = − + − + =
4EF?G>HC5
*
&
- *x x x x x− − + + − = =
!*
& . - *x x x x x− + − = − + =
*
)
I
x x x x
− ±
+ = − = =
÷
÷
*
+
I
x x x x
+ + − = = ± = −
÷
*
, , - & .*x x x x x− − + = + = ±
J*
. ) , ' - *x x x− + − − = = −
5EF?G>HC5
*
- *
x x
x
x
x
+ −
+ = =
−
!*
& & - ,*x x x x x− + − − = =
*
)
) ' &
x x x x x
±
+ + − + + − = =
÷
÷
*
) . - &*x x x+ + + = =
*
( & - *x x x− + + = =
J*
& ) - &*x x x+ + − = =
*
' - *x x x x− + − + = =
*
- I )*
x
x x x x x x
+
+ − + − − = = =
6EF/G>HC5
*
x x x x+ + + = + +
-KL*
!*
x
x x x x
x
+
+ + = − + + +
+
-
7 x x= − = +
*
*
x x x x+ + + = + + +
-KL'IKLM*
*
x x x x x+ + = + +
-KL*
*
&
&
x
x x
x
+ + =
+
-KL*
I. C ơ bản :
E
( &x x x+ = − −
E
x x x x+ + + = + +
E
- * .x x x+ − = + +
&E
x x x x+ + + = + + +
)E
x x x x x+ + = + +
.E
& x x x x x x+ + + = + + +
+E
- *x x x x x x− − − − + −
,E
, . x x x x+ + + − = +
(E
) 'x x x− − − − − =
'E
( )
' x x x x+ − = − −
E
& x x x x− − + + − − =
E
)
x
x x x x
+
+ + + + + − + =
II. È n phu :
E
x x x x− − + + − =
&E
=++
+
xx
x
)E
) .x x+ + − =
.E
& &x x x x+ − = + −
+E
x x x x+ − = + −
,E
x x x x
x
+ − = +
(E
&
x x x x+ − = +
'E
+
− + + − = −
−
1
( 3)( 1) 4( 3) 3
3
x
x x x
x
E
& ( ) x x x x x
− + − = − + − +
E
x x+ + =
E
( )
) x x+ = +
&E
2 3
2 5 1 7 1x x x+ − = −
)E
− + = + +
2 2
(4 1) 1 2 2 1x x x x
.E
2 2
2(1 ) 2 1 2 1x x x x x− + − = − −
+E
( )
. 'x x x x− + + − =
,E
& x x x x x+ + − = + +
(E
(
)
x x x x+ − + = + +
'E
& x x x x+ − = + − + −
E
-& * x x x x− + = + +
E
- * x x x x+ + = + +
E
+ +x x+ + =
&E
+ = −
3
3
1 2 2 1x x
)E
x x+ = −
.E
. , & x x x+ = − −
+E
x x− + − =
,E
2 2
3 2 1( 99)x x x x NT− + − + − = −
(E
x x
− = − −
&'E
( &x x x x x+ + + − + = +
&E
x x x− = −
&E
. & )x x x− − = +
&E
(
)
x x x+ − = + −
&&E
( ) ( )
x x x x+ − = −
&)E
( ) ( )
x x x x
+ − − − + = + −
&.E
. x x+ =
&
)