đề tài phương trình vô tỷ

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LỜI NÓI ĐẦU
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
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I. PHƯƠNG PHÁP BIẾN ĐỔI TƯƠNG ĐƯƠNG
1. Bình phương 2 vế của phương trình
a) Phương pháp
 7@:;<78U:;:vT<N;r__789:;<=>:7YO:; 
A B C D+ = +
R<N<78U:;
k>:7_789:;3?vRFHGTFlF@Hs7HBOH;r_s7ls7•:
 
( )
Œ Œ Œ Œ
Œ Œ
Œ `A B C A B A B A B C+ = ⇒ + + + =

?C<NP•Yn:;_7Ž_<7v
Œ Œ
A B C+ =
<NF8aL_789:;<=>:7 
Œ

Œ ` `A B A B C C+ + =
b) Ví dụ
B,i 1.H]H_789:;<=>:7PNT 
3 Œ 1x x− + =
•0•
"#"$s
1x ≥

3
3
•0• 3 Œ
3 Œ
3 Œ 1
Œ
x x
x x
x x
x
⇔ = +
⇔ = +
⇔ − − =
⇔ =
ˆX<ˆ_:;7HpDLMN_789:;<=>:7BC
{ }
ŒS =
B,i 2.H]H_789:;<=>:7PNT 
"#"$ 2. Trục căn thức
2.1. Trục căn thức để xuất hiện nhân tử chung
a) Phương pháp
yHDE<PQ_789:;<=>:7<NLl<7\:7‘DF8aL:;7HpD

1
x
:78?ˆX_789:;<=>:7
BT@:F8N?GF8aLYO:;<uL7
( )
( )
1
1x x A x− =
<NLl<7\;H]H_789:;<=>:7
( )
1A x =

7[rLL7W:;DH:7
( )
1A x =
?@:;7HpDR%&'()"*+," %/0.1&"-2%/03&45.1678.&
)960%:6&9);.&1<0
( )
1A x =
=>.1&"-2
b) Ví dụ
B,i 1 .H]H_789:;<=>:7PNT 
( )
3 3 3 3
Œ ’ 0 3 Œ 0 Œ “x x x x x x x− + − − = − − − − +
"#"$
N:7ˆ:<7cX 
( ) ( )
( )
3 3

Œ ’ 0 Œ Œ Œ 3 3x x x x x− + − − − = − −


( ) ( )
( )
3 3
3 Œ “ Œ 3x x x x
− − − + = −
NLl<7\L7TX\:?v=…H<=nLL•:<7WL3?v
( )
3 3
3 3
3 “ Œ ”
3 Œ “
Œ ’ 0 Œ 0
x x
x x x
x x x x
− + −
=
− + − +
− + + − +
\YC:;:7ˆ:<7cX‹•3BC:;7HpDYTX:7c<LMN_789:;<=>:7`
B,i 2.H]H_789:;<=>:7PNT(OLYMPIC 30/4 đề nghị) 
3 3
03 ’ Œ ’x x x+ + = + +
"#"$\_789:;<=>:7Ll:;7HpD<7> 
3 3
’
03 ’ Œ ’ 1

Œ
x x x x+ − + = − ≥ ⇔ ≥
N:7ˆ:<7cX ‹•3BC:;7HpDLMN_789:;<=>:7R:78?ˆX_789:;<=>:7Ll<7\_7d:<uL7
?GYO:;
( ) ( )
3 1x A x− =
RF\<7zL7Hp:F8aLFHGTFl<N_7]H:7lDR<ZL7LZLPQ7O:;:78PNT
( )
( )
3 3
3 3
3 3
3 3
“ “
03 “ Œ ” ’ Œ Œ 3
03 “ ’ Œ
3 0
3 Œ 1
03 “ ’ Œ
3
x x
x x x x
x x
x x
x
x x
x
− −
+ − = − + + − ⇔ = − +
+ + + +

 
+ +
⇔ − − − =
 ÷
+ + + +
 
⇔ =
–YC:;L7W:;DH:7F8aL 
3 3
3 3 ’
Œ 1R
Œ
03 “ ’ Œ
x x
x
x x
+ +
− − < ∀ >
+ + + +
B,i 3.H]H_789:;<=>:7
3 ŒŒ
0 3x x x− + = −
"#"$s
Œ
3x ≥
7ˆ:<7cX‹•ŒBC:;7HpDLMN_789:;<=>:7R:V:<NkHv:F•H_789:;<=>:7
( )
( )
( )
( )

3
3 ŒŒ
3 Œ
3 3Œ
Œ
Œ Œ 6
Œ
0 3 Œ 3 ’ Œ 0
3 ’
0 3 0 “
x x x
x
x x x x
x
x x
 
− + +
+
 
− − + − = − − ⇔ − + =
 
− +
− + − +
 
 
NL7W:;DH:7
( )
(
)
3

3
3 Œ
3 3 3
Œ Œ
Œ
Œ Œ Œ 6
0 0 3
3 ’
0 3 0 “ 0 0 Œ
x x x x
x
x x x
+ + + +
+ = + < <
− +
− + − + − + +
ˆX_<Ll:;7HpDYTX:7c<‹•Œ
2.2. Đưa về “hệ tạm “
a) Phương pháp
 vT_789:;<=>:7?@<‚LlYO:;
A B C+ =
RDC 
A B C
α
− =

^YdXLl<7\BC7C:;PQRLl<7\BCkH\T<7WLLMN
x
`NLl<7\;H]H:78PNT
A B

C A B
A B
α
−
= ⇒ − =
−
Rs7HFl<NLl7p 
3
A B C
A C
A B
α
α

+ =

⇒ = +

− =


b) Ví dụ
B,i 4. H]H_789:;<=>:7PNT
3 3
3 6 3 0 “x x x x x+ + + − + = +
Giải:
N<7cX 
( ) ( )
( )
3 3

3 6 3 0 3 “x x x x x+ + − − + = +
“x = −
s7@:;_7]HBC:;7HpD
—Ž<
“x
> −
=nLL•:<7WL<NLl
3 3
3 3
3 4
“ 3 6 3 0 3
3 6 3 0
x
x x x x x
x x x x
+
= + ⇒ + + − − + =
+ + − − +
ˆX<NLl7p
3 3
3
3 3
1
3 6 3 0 3
3 3 6 ”
4
3 6 3 0 “
˜
x
x x x x

x x x
x
x x x x x
=


+ + − − + =


⇒ + + = + ⇔


=
+ + + − + = +



ˆX_789:;<=>:7Ll3:;7HpD ‹•1?‹•
4
˜
B,i 5.H]H_789:;<=>:7 
3 3
3 0 0 Œx x x x x+ + + − + =
N<7cX 
( ) ( )
3 3 3
3 0 0 3x x x x x x+ + − − + = +
R•s7@:;LlYcT7HpT<=V:•`
NLl<7\L7HNL]7NH?vL7[‹?CFr<
0

t
x
=
<7>kCH<[Z:<=^:V:F9:;H]:79:
3. Phương trình biến đổi về tích
 Sử dụng các đẳng thức

( ) ( )
0 0 0 1u v uv u v+ = + ⇔ − − =

( ) ( )
1au bv ab vu u b v a+ = + ⇔ − − =

3 3
A B=
B,i 1. H]H_789:;<=>:7 
3Œ
Œ Œ
0 3 0 Œ 3x x x x+ + + = + + +
"#"$
( ) ( )
Œ Œ
1
0 0 3 0 1
0
x
pt x x
x
=


⇔ + − + − = ⇔

= −

B,i 2. H]H_789:;<=>:7 
3 3Œ Œ
Œ Œ
0x x x x x+ + = + +
"#"$
™
1x =
Rs7@:;_7]HBC:;7HpD
™
1x
≠
R<NL7HN7NH?vL7[‹
( )
Œ Œ Œ
Œ Œ
0 0
0 0 0 0 1 0
x x
x x x x
x x
 
+ +
+ = + + ⇔ − − = ⇔ =
 ÷
 
B,i 3. H]H_789:;<=>:7 

3
Œ 3 0 3 “ Œx x x x x x+ + + = + + +
"#"$s
0x ≥ −
_<
( ) ( )
0
Œ 3 0 0 1
1
x
x x x
x
=

⇔ + − + − = ⇔

=

B,i 4.H]H_789:;<=>:7 
“
Œ “
Œ
x
x x
x
+ + =
+
Giải:
s 
1x

≥
7HNL]7NH?vL7[
Œx +

3
“ “ “
0 3 0 1 0
Œ Œ Œ
x x x
x
x x x
 
+ = ⇔ − = ⇔ =
 ÷
+ + +
 
 Dùng hằng đẳng thức
/Hv:F•H_789:;<=>:7?GYO:;
k k
A B=
B,i 1.H]H_789:;<=>:7 
Œ Œx x x− = +
"#"$
s 
1 Œx≤ ≤
s7HFl_<FSL7[<89:;F89:; 
Œ 3
Œ Œ 1x x x+ + − =
Œ
Œ

0 01 01 0
Œ Œ Œ Œ
x x
−
 
⇔ + = ⇔ =
 ÷
 
B,i 2.H]H_789:;<=>:7PNT
3
3 Œ 6 “x x x+ = − −
"#"$
s
Œx ≥ −
_789:;<=>:7<89:;F89:;
( )
3
3
0
Œ 0 Œ
0 Œ 6
’ 6˜
Œ 0 Œ
04
x
x x
x x
x
x x
=



+ + =

+ + = ⇔ ⇔

− −

=
+ + = −




B,i 3.H]H_789:;<=>:7PNT 
( ) ( )
3
3
Œ
Œ
3 Œ 6 3 3 Œ Œ 3x x x x x+ + = + +
"#"$_<
( )
Œ
Œ Œ
3 Œ 1 0x x x⇔ + − = ⇔ =
B,i tập đề nghị
H]HLZL_789:;<=>:7PNT
1)
( )

3 3
Œ 0 Œ 0x x x x+ + = + +
2)
“ Œ 01 Œ 3x x− − = −
?@AB.C+D%EEF
3)
( ) ( ) ( ) ( )
3 3 ’ 3 01x x x x x− − = + − −
4)
3Œ
“ 0 3 Œx x x+ = − + −
5)
3 ŒŒ
0 Œ 3 Œ 3x x x− + − = −
6)
3
Œ
3 00 30 Œ “ “ 1x x x− + − − =
?GHIEJKEEF
7)
3 3 3 3
3 0 Œ 3 3 3 Œ 3x x x x x x x− + − − = + + + − +
8)
3 3
3 0” 04 0 3 “x x x x+ + + − = +
9)
3 3
0’ Œ 3 4x x x+ = − + +
II. PHƯƠNG PHÁP ĐẶT ẦN PHỤ
1. Phương pháp đặt ẩn phụ thông thường

 QH?yH:7HGT_789:;<=>:7?@<‚RF\;H]HL7J:;<NLl<7\Fr<
( )
t f x=
?CL7J
IFHGTsHp:LMN
t
:vT_789:;<=>:7kN:FxT<=^<7C:7_789:;<=>:7L7WNDE<kHv:
t
wTN:<=h:;79:<NLl<7\;H]HF8aL_789:;<=>:7Fl<7†[
t
<7>?HpLFr<_7n‹†D
:78|7[C:<[C:~`lHL7T:;:7g:;_789:;<=>:7DCLl<7\Fr<7[C:<[C:
( )
t f x=
<78U:;BC:7g:;_789:;<=>:7Y–`
B"LH]H_789:;<=>:7 
3 3
0 0 3x x x x− − + + − =
"#"$
s 
0x ≥
7ˆ:‹Ž<`
3 3
0` 0 0x x x x− − + − =
r<
3
0t x x= − −
<7>_789:;<=>:7LlYO:; 
0
3 0t t

t
+ = ⇔ =
7NX?C[<>DF8aL
0x
=
B"H]H_789:;<=>:7 
3
3 ” 0 “ ’x x x− − = +
"#"
HGTsHp: 
“
’
x ≥ −
r<
“ ’• 1•t x t= + ≥
<7>
3
’
“
t
x
−
=
`7NX?C[<NLl_789:;<=>:7PNT
“ 3
3 “ 3
01 3’ ”
3` • ’• 0 33 4 3˜ 1
0” “
t t

t t t t t
− +
− − − = ⇔ − − + =
3 3
• 3 ˜•• 3 00• 1t t t t⇔ + − − − =
N<>DF8aLkQ::;7HpDBC 
0R3 ŒR“
0 3 3„ 0 3 Œt t= − ± = ±
[
1t ≥
:V:L7‚:7ˆ:LZL;ZH<=K
0 Œ
0 3 3R 0 3 Œt t= − + = +
bFl<>DF8aLLZL:;7HpDLMN_789:;<=>:7B 
0 3 3 Œ vaø x x= − = +
Cách khác: NLl<7\k>:7_789:;7NH?vLMN_789:;<=>:7?yHFHGTsHp:
3
3 ” 0 1x x− − ≥
NF8aL 
3 3 3
• Œ• • 0• 1x x x− − − =
R<bFl<N<>DF8aL:;7HpD<89:;W:;`
9:;H]::7c<BC<NFr< 
3 Œ “ ’y x− = +
?CF8N?G7pFQH‹W:;
•Xem phần dặt ẩn phụ đưa về hệ)
B"H]H_789:;<=>:7PNT 
’ 0 ”x x+ + − =
HGTsHp: 
0 ”x

≤ ≤
r<
0• 1•y x y= − ≥
<7>_789:;<=>:7<=^<7C:7
3 “ 3
’ ’ 01 31 1y y y y y+ + = ⇔ − − + =
•?yH
’•y ≤
3 3
• “•• ’• 1y y y y⇔ + − − − =
0 30 0 0˜
R
3 3
(loaïi)y y
+ − +
⇔ = =
bFl<N<>DF8aLLZL;HZ<=KLMN
00 0˜
3
x
−
=
B,i 4`(THTT 3-2005) H]H_789:;<=>:7PNT
( )
(
)
3
311“ 0 0x x x= + − −
Giải:Fs
1 0x≤ ≤

r<
0y x= −
_<<<
( )
( )
3
3
3 0 0113 1 0 1y y y y x
⇔ − + − = ⇔ = ⇔ =
B,i 5.H]H_789:;<=>:7PNT 
3
0
3 Œ 0x x x x
x
+ − = +
"#"$
HGTsHp: 
0 1x
− ≤ <
7HNL]7NH?vL7[‹<N:7ˆ:F8aL
0 0
3 Œx x
x x
+ − = +
r<
0
t x
x
= −
R<N;H]HF8aL`

B,i 6.H]H_789:;<=>:7 
3 “ 3Œ
3 0x x x x+ − = +
H]H 
1x =
s7@:;_7]HBC:;7HpDR7HNL]7NH?vL7[‹<NF8aL
Œ
0 0
3x x
x x
 
− + − =
 ÷
 
r<<•
Œ
0
x
x
−
RNLl 
Œ
3 1t t+ − = ⇔
0 ’
0
3
t x
±
= ⇔ =
&M.NO6 FQH?yHLZL7Fr<‘:_7n:78<=V:L7J:;<NL7‚;H]HwTXv<F8aLDE<By_

kCHF9:;H]:RF@Hs7H_789:;<=>:7FQH?yH
t
BOHwTZs7l;H]H
2. Đặt ẩn phụ đưa về phương trình thuần nhất bậc 2 đối với 2 biến :
 7J:;<NFSkHv<LZL7;H]H_789:;<=>:7 
3 3
1u uv v
α β
+ + =
•0•kš:;LZL7
—Ž<
1v ≠
_789:;<=>:7<=^<7C:7 
3
1
u u
v v
α β
   
+ + =
 ÷  ÷
   
1v
=
<7•<=zL<Hv_
ZL<=8U:;7a_PNTL›:;F8N?GF8aL•0•

( ) ( ) ( ) ( )
` `a A x bB x c A x B x+ =


3 3
u v mu nv
α β
+ = +
7J:;<N7SX<7NXLZLkH\T<7WL•‹•R/•‹•k^HLZLkH\T<7WL?@<‚<7>P‡:7ˆ:
F8aL_789:;<=>:7?@<‚<7†[YO:;:CX`
a) Phương trình dạng :
( ) ( ) ( ) ( )
` `a A x bB x c A x B x+ =
78?ˆX_789:;<=>:7
( ) ( )
Q x P x
α
=
Ll<7\;H]Hkš:;_789:;_7Z_<=V::vT

( ) ( ) ( )
( ) ( ) ( )
`P x A x B x
Q x aA x bB x
 =


= +


—Tc<_7Z<<bFœ:;<7WL

( )
( )

Œ 3
0 0 0x x x x+ = + − +

( ) ( ) ( )
“ 3 “ 3 3 3 3
0 3 0 0 0x x x x x x x x x+ + = + + − = + + − +

( ) ( )
“ 3 3
0 3 0 3 0x x x x x+ = − + + +

( ) ( )
“ 3 3
“ 0 3 3 0 3 3 0x x x x x+ = − + + +
SX<O[=N:7g:;_789:;<=>:7?@<‚YO:;<=V:?uYn:78 

3 “
“ 3 3 “ 0x x x− + = +
\LlDE<_789:;<=>:7F‰_RL7J:;<N_7]HL7h:7pPQNRkRLPN[L7[_789:;<=>:7
kˆL7NH
3
1at bt c+ − =
;H]H|:;7HpDF‰_~
B,i 1. H]H_789:;<=>:7 
( )
3 Œ
3 3 ’ 0x x+ = +
Giải:r<
3
0R 0u x v x x= + = − +


_789:;<=>:7<=^<7C:7 
( )
3 3
3
3 ’
0
3
u v
u v uv
u v
=


+ = ⇔

=

>DF8aL 
’ Œ˜
3
x
±
=
B,i 2.H]H_789:;<=>:7
3 “ 3
Œ
Œ 0 0
Œ
x x x x− + = − + +

B,i 3:;H]H_789:;<=>:7PNT
3 Œ
3 ’ 0 ˜ 0x x x+ − = −
"#"$
s 
0x ≥
7ˆ:‹Ž< N?Hv<
( )
( )
( )
( )
3 3
0 0 ˜ 0 0x x x x x x
α β
− + + + = − + +
…:;:7c<<7WL<NF8aL
( ) ( ) ( )
( )
3
Œ 0 3 0 ˜ 0 0x x x x x x− + + + = − + +
r<
3
0 1 R 0 1u x v x x= − ≥ = + + >
R<NF8aL 
6
Œ 3 ˜
0
“
v u
u v uv

v u
=


+ = ⇔

=

;7HpD
“ ”x = ±
B,i 4.H]H_789:;<=>:7
( )
Œ
Œ 3
Œ 3 3 ” 1x x x x− + + − =
H]H
7ˆ:‹Ž< r<
3y x= +
<NkHv:_<<=>:7?GYO:;_789:;<=>:7<7Tx::7c<kˆLŒ
FQH?yH‹?CX
Œ 3 Œ Œ 3 Œ
Œ 3 ” 1 Œ 3 1
3
x y
x x y x x xy y
x y
=

− + − = ⇔ − + = ⇔


= −

<Ll:;7HpD
3R 3 3 Œx x= = −
b).Phương trình dạng :
3 3
u v mu nv
α β
+ = +
789:;<=>:7L7[^YO:;:CX<78U:;s7l|_7Z<7Hp:|79:YO:;<=V:R:78:;
:vT<Nk>:7_789:;7NH?v<7>F8N?GF8aLYO:;<=V:`
B,i 1. H]H_789:;<=>:7 
3 3 “ 3
Œ 0 0x x x x+ − = − +
Giải:
NFr<
3
3
0
u x
v x

=


= −


s7HFl_789:;<=>:7<=^<7C:7 
3 3

Œu v u v+ = −
B,i 2.H]H_789:;<=>:7PNT 
3 3
3 3 0 Œ “ 0x x x x x+ + − = + +
H]H
s
0
3
x ≥
`/>:7_789:;3?v<NLl
( )
( )
( )
( )
( )
( )
3 3 3 3
3 3 0 0 3 3 0 3 3 0x x x x x x x x x x+ − = + ⇔ + − = + − −
NLl<7\Fr< 
3
3
3 0
u x x
v x

= +

= −

s7HFl<NLl7p 

3 3
0 ’
3
0 ’
3
u v
uv u v
u v

−
=


= − ⇔

+
=


[
R 1u v ≥
`
( )
3
0 ’ 0 ’
3 3 0
3 3
u v x x x
+ +
= ⇔ + = −

B,i 3.;H]H_789:;<=>:7 
3 3
’ 0“ 6 31 ’ 0x x x x x− + − − − = +
H]H
s
’x
≥
`7TX\:?vk>:7_789:;<NF8aL 
( )
( )
3 3
3 ’ 3 ’ 31 0x x x x x− + = − − +
Nhận xét : s7@:;<…:<OHPQ
R
α β
F\ 
( )
( )
3 3
3 ’ 3 31 0x x x x x
α β
− + = − − + +

?ˆX<Ns7@:;<7\Fr<
3
31
0
u x x
v x


= − −

= +

`
78:;DNXD{:<NLl
( )
( ) ( ) ( ) ( ) ( )
( )
3 3
31 0 “ ’ 0 “ “ ’x x x x x x x x x− − + = + − + = + − −
N?Hv<BOH_789:;<=>:7 
( )
( )
3 3
3 “ ’ Œ “ ’ • “ ’•• “•x x x x x x− − + + = − − +
`v:
FdXkCH<[Z:F8aL;H]HwTXv<`
Các bạn hãy tự tạo cho mình những phương trình vô tỉ “đẹp “ theo cách trên
3. Phương pháp đặt ẩn phụ không ho,n to,n
 b:7g:;_789:;<=>:7<uL7
( ) ( )
0 0 0 3 1x x x+ − + − + =
R
( ) ( )
3 Œ 3 Œ 3 1x x x x+ − + − + =
#7NH<=H\:?C=J<;h:<NP‡F8aL:7g:;_789:;<=>:7?@<‚s7@:;<xD<78U:;L7J<
:C[RFEs7lLMN_789:;<=>:7YO:;:CX_7n<7TEL?C[_789:;<=>:7<uL7DC<N‹Tc<
_7Z<`
bFlL7J:;<NDyHFH<>DLZL7;H]H_789:;<=>:7YO:;:CX`789:;_7Z_;H]H

F8aL<7\7Hp:wTNLZL?uYnPNT`
B,i 1.H]H_789:;<=>:7
(
)
3 3 3
Œ 3 0 3 3x x x x+ − + = + +
"#"$
3
3t x= +
R<NLl 
( )
3
Œ
3 Œ Œ 1
0
t
t x t x
t x
=

− + − + = ⇔

= −

B,i 2`H]H_789:;<=>:7 
( )
3 3
0 3 Œ 0x x x x+ − + = +
"#"$
r< 

3
3 ŒR 3t x x t= − + ≥
#7HFl_789:;<=>:7<=^<7C:7 
( )
3
0 0x t x+ = +
( )
3
0 0 1x x t⇔ + − + =
/dX;HU<N<7VDky<RF\F8aL_789:;<=>:7kˆL3<7†[<
( ) ( ) ( ) ( )
3 3
3
3 Œ 0 3 0 1 0 3 0 1
0
t
x x x t x t x t x
t x
=

− + − + + − = ⇔ − + + − = ⇔

= −

bDE<_789:;<=>:7F9:;H]: 
( ) ( )
0 3 0 0 3 0 1x x x x− − + − − + + =
Rs7NH
<=H\:=N<NP‡F8aL_<PNT
B,i 3`H]H_789:;<=>:7PNT 

3
“ 0 0 Œ 3 0 0x x x x+ − = + − + −
"#"$
7ˆ:‹Ž< Fr<
0t x= −
R_<<=^<7C:7
“ 0 Œ 3 0x x t t x+ = + + +
•0•
N=J<
3
0x t= −
<7NX?C[<7>F8aL_< 
( ) ( )
3
Œ 3 0 “ 0 0 1t x t x− + + + + − =
78:;s7@:;LlPzDNXD{:F\;H]HF8aL_789:;<=>:7<7†[<
( ) ( )
3
3 0 “4 0 0x x∆ = + + − + −
s7@:;LlYO:;k>:7_789:;`
$TQ:FO<F8aLDnLFuL7<=V:<7><N_7]H<ZL7Œ‹<7†[
( ) ( )
3 3
0 R 0x x− +
n<7\:78PNT 
( ) ( )
Œ 0 3 0x x x= − − + +
<7NX?C[_<•0•
B,i 4`H]H_789:;<=>:7 
3

3 3 “ “ 3 6 0”x x x+ + − = +
"#"$
/>:7_789:;3?v_789:;<=>:7 
( )
( )
( )
3 3
“ 3 “ 0” 3 “ 0” 3 6 0”x x x x+ + − + − = +
NFr< 
( )
3
3 “ 1t x= − ≥
`NF8aL 
3
6 0” Œ3 4 1x t x− − + =
N_7]H<ZL7
( )
( )
3 3 3
6 3 “ 6 3 4x x x
α α α
= − + + −
BCDPN[L7[
t
∆
LlYO:;PQL7u:7
_789:;`
&M.NO6$7@:;<78U:;<NL7‚Lx::7lDPN[L7[7v<7pPQ<zY[<7>P‡FO<
F8aLDnLFuL7
4. Đặt nhiều ẩn phụ đưa về tích

 —Tc<_7Z<<bDE<PQ7p|FOHPQ|F‰_L7J:;<NLl<7\<O[=NF8aL:7g:;
_789:;<=>:7?@<‚DCs7H;HCH:lL7J:;<NBOHFr<:7HGT‘:_7n?C<>DDQHwTN:
7p;HgNLZL‘:_7nF\F8N?G7p
—Tc<_7Z<<bFœ:;<7WL
( ) ( ) ( ) ( )
Œ
Œ Œ Œ
Œa b c a b c a b b c c a+ + = + + + + + +
RNLl
( ) ( ) ( ) ( )
Œ
Œ Œ Œ
1a b c a b c a b a c b c+ + = + + ⇔ + + + =
b:7ˆ:‹Ž<:CX<NLl<7\<O[=N:7g:;_789:;<=>:7?@<‚LlL7WNL•:kˆLkN`
3 3Œ Œ
Œ
˜ 0 4 4 0 3x x x x x+ − − − + − + =
Œ Œ Œ Œ
Œ 0 ’ 3 6 “ Œ 1x x x x+ + − + − − − =
B,i 1. H]H_789:;<=>:7
3 ` Œ Œ ` ’ ’ ` 3x x x x x x x= − − + − − + − −
H]H 
3
Œ
’
u x
v x
w x

= −



= −


= −


R<NLl 
( ) ( )
( ) ( )
( ) ( )
3
3
3
3
3
Œ Œ
’
’
u v u w
u uv vw wu
v uv vw wu u v v w
w uv vw wu
v w u w
 + + =

− = + +



− = + + ⇔ + + =
 
 
− = + +
+ + =
 
R;H]H7p
<NF8aL 
Œ1 3Œ6
”1 031
u x= ⇔ =
B,i 2.H]H_789:;<=>:7PNT 

3 3 3 3
3 0 Œ 3 3 3 Œ 3x x x x x x x− + − − = + + + − +
"#"NFr< 
3
3
3
3
3 0
Œ 3
3 3 Œ
3
a x
b x x
c x x
d x x

= −



= − −


= + +


= − +


Rs7HFl<NLl 
3 3 3 3
3
a b c d
x
a b c d
+ = +

⇔ = −

− = −

B,i 3. H]HLZL_789:;<=>:7PNT
0•
3 3
“ ’ 0 3 0 6 Œx x x x x+ + − − + = −
3•
( ) ( ) ( )
Œ

Œ 3
“
“
“
“
0 0 0 0x x x x x x x x+ − + − = − + + −
B,i tập đề nghị
Giải các phương trình sau
N`
3 3
0’ 3 ’ 3 0’ 00x x x x− − = − +
k`
3
• ’••3 • Œ Œx x x x+ − = +
L`
3
•0 ••3 • 0 3 3x x x x+ − = + −
Y`
3 3
0˜ 0˜ 6x x x x+ − + − =
†`
3
Œ 3 0 “ 6 3 Œ ’ 3x x x x x− + − = − + − +
•`
3 3
00 Œ0x x+ + =
;`
3 3 3
3 •0 • Œ 0 •0 • 1
n

n n
x x x+ + − + − =
7`
3
•311“ ••0 0 •x x x= + − −
H`
• Œ 3•• 6 04• 0”4x x x x x+ + + + =
ž`
Œ
3 3
0 3 0 Œx x− + − =
B,i tập tổng hợp:
0ƒ
nxxxx =−+−−++ •Œ••0•Œ0
•0•
NƒH]H_789:;<=>:7:•3
kƒ>DLZL;HZ<=KLMN:F\_789:;<=>:7Ll:;7HpD 
3ƒ
”
6”6”
mx
xxxx
+
=−−+−+
NƒH]H_789:;<=>:7?yHD•3Œ
kƒ>DLZL;HZ<=KLMNDF\_789:;<=>:7Ll:;7HpD`
/CH<ˆ_<89:;<z
Œƒ
30303 =−−+−+ xxxx
“ƒ

xxxx −+=−+ 0
Œ
3
0
3
’ƒ
0”Œ’33Œ0Œ3
3
−+++=+++ xxxxx
”ƒ
3
’0013 xxx =+−−
˜ƒ>DDF\R_789:;<=>:7PNTLl:;7HpD
mxxxx =−++−++ •4••0•40
4ƒ>DDF\_789:;<=>:7PNTLl:;7HpD 
mxxxx =−−+−−− “““3Œ
6ƒ>DDF\_789:;<=>:7PNTLl:;7HpD
mxxxx =−++−+ “““
H]HLZL_789:;<=>:7PNT
01ƒ
”0”
3
““
3
−−+=
−++
xx
xx
00ƒ
“3330””Œ

333
+=++++ xxxxxx
03ƒ
˜Œ33Œ•Œ•
33
+−=−+− xxxx
0Œƒ
101““Œ
33
=+−+− xxxx
0“ƒ
10Œ“3“•3•Œ
33
=++−−− xxxx
0’ƒ
0•03••0•030 −=−++−++ xxxx
0”ƒ
0•4••’•4’ −=−++−++ xxxx
0˜ƒ
•0•3˜003“
3
xxxx +=++
04ƒ
•0•0103
3
−=−+ xxxx
06ƒ
10’0003“’03“
33
=++−−− xxx

31ƒ
03
Œ’
0
3
=
−
+
x
x
x
•0•
30ƒ
Œ
3
Œ
33
Œ
3
Œ
3
•••••• aaxaxax =−++++
33ƒ
4˜˜”36
““
=++− xx
3Œƒ
( )
3
Œ

Œ3
030•0•00 xxxx −+=






−−+−+
3“ƒ
xxx −
=
−+
+
−− 0
Œ“
00
0
00
0
3’ƒ
0”Œ’33Œ0Œ3
3
−+++=+++ xxxxx
3’ƒ
0”Œ’33Œ0Œ3
3
−+++=+++ xxxxx
3”ƒ>D<c<L]LZL;HZ<=K<7zLLMN<7NDPQ
α

F\_789:;<=>:7PNTLl
:;7HpD
•0••3•
Œ
0
•Œ•Œ•0••Œ• −+=
−
+
−++− aa
x
x
xxx
/CH<ˆ_<89:;<z
3˜ƒ
•3••Œ•
3
“
•3•’•“••3• +−=
+
+
++++ aa
x
x
xxx
34ƒ
”
•3’•
0”
•3’•
’

Œ
’
Œ
=
+
−+
x
x
36ƒ
3
’
0”
0
0
0”
’’
=
−
+
− y
y
y
y
Œ1ƒ
“
’
“
’
””
=

+
+
+
x
x
x
x
Œ0ƒ
3
’
Œ
Œ
’
˜˜
=
−
+
+
+
−
x
x
x
x
Œ3ƒ
33
31
=++ xxx
x
•r<X•

x
Ÿ1•
ŒŒƒ
•3•3•3’•
3
’
Œ
+=+ xx
•0•
Œ“ƒ
Œ30
3Œ3Œ
=+−+−− xxxx
•0•
Œ’ƒ
3
3
Œ
3
0






−=− xx
Œ”ƒ
“’”“’04
““

=−++ xx
Œ˜ƒ
’’’
3axaxa =−++
Œ4ƒ
00303003 ++=+−++++ xxxxx
Œ6ƒ
Œ30
3Œ3Œ
=+++−+ xxxx
“0ƒ
344
““
=−−+ xx
“3ƒ
3
3
0
3
0
3
=
−
+
x
GIẢI BÀI TẬP TỔNG HỢP
0ƒ
nxxxx =−+−−++ •Œ••0•Œ0
•0•
NƒH]H_789:;<=>:7:•3

kƒ>DLZL;HZ<=KLMN:F\_789:;<=>:7Ll:;7HpD 
HGTsHp:



≤≤−⇔
≥−
≥+
Œ0
1Œ
10
x
x
x
r<‘:_7n
1RŒ0 ≥−++= txxt
#7HFl
•Œ••0•3“
3
xxt −++=
NX
“•Œ••0•3
3
−=−+ txx
•3•
NƒyH:•3?C‘:_7n<R_789:;<=>:7•0•<=^<7C:7`
3R1
13
“•“•3
30

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