Bµi tËp hÖ ph¬ng tr×nh
Gi¶i c¸c hÖ ph¬ng tr×nh sau :
1,
+ + = −
−
+ = −
2 2
1
( 99)
6
x xy y
MTCN
x y y x
2,
+ =
−
− + =
2 2
4 2 2 4
5
( 98)
13
x y
1
( 97)
x y
AN
x y x y
5,
+ + =
−
+ + =
2 2
4 4 2 2
7
( 1 2000)
21
x y xy
SP
x y x y
6,
+ + =
−
+ + + =
2 2
+ + =
2 2
2 2
1
( )(1 ) 5
( 99)
1
( )(1 ) 49
x y
xy
NT
x y
x y
9,
+ + + =
−
+ + + =
2 2
2 2
+ + + − + + + + − =
2 2
2 2
1 1 18
( 99)
1 1 2
x x y x y x y y
AN
x x y x y x y y
12,
+ + =
−
+ + − =
2
(3 2 )( 1) 12
( 97)
2 4 8 0
x x y x
BCVT
x y x
13,
+ =
−
−
− = −
2 2
2 2
2 3 2
( 2000)
2 3 2
x x y
QG
y y x
16,
= −
−
= −
2
2
3
( 98)
3
x x y
MTCN
y y x
3
3
3 8
( 98)
3 8
x x y
QG
y y x
19,
+ =
−
+ =
2
2
3
2
( 2001)
3
2
x y
x
TL
y x
2
2
2
2
3
( 2003)
2
3
y
y
x
KhèiB
x
x
y
22,
− =
−
− − =
2
2 2
3 2 16
( )
3 2 8
x xy
2 3 9
( )
2 13 15 0
x xy y
HVNH TPHCM
x xy y
25,
− =
−
+ =
2 2
2 2
2 ( ) 3
( § 97)
( ) 10
y x y x
M C
x x y y
Bài tập phơng trình -bất phơng trình vô tỉ
Giải các phơng trình sau:
1,
3 6 3x x+ + =
2,
9 5 2 4x x+ = +
3,
2 8 6 1 2 2( 99)x x x x HVKTQS+ + + = +
Tìm m để phơng trình :
14,
2
2 2 1( 2006)x mx x KhốiB+ + = +
có 2 nghiệm phân biệt
15,
2
2 3 ( )x mx x SPKT TPHCM+ =
có nghiệm
16,
2
2 3 ( 98)x mx x m GT+ =
có nghiệm
Giải các phơng trình sau :
17,
2 2
11 31x x+ + =
18,
2
( 5)(2 ) 3 3x x x x+ = +
19,
2 2
3 3 3 6 3( 98)x x x x TM + + + =
20,
2 3
2 5 1 7 1x x x+ =
21,
2 3
2 4 3 4x x x x+ + = +
2 1 1
2( 95)
1 2 2
x
GT
x x
+ + =
+
30,
2
2 2
1
x
x
x
+ =
31,
2 2
1 1 (1 2 1 )x x x+ = +
32,
2 2
(4 1) 1 2 2 1(Đ 78)x x x x ề + = + +
33,
2 2
3 1 ( 3) 1( 01)x x x x GT+ + = + +
34,
2 2
2(1 ) 2 1 2 1x x x x x + =
35,
3 2 8 7 ( 97)x x x AN+ ≥ − + − −
4,
2 3 5 2 ( 2000)x x x TL+ − − < − −
5,
2 2
( 3) 4 9(§ 11)x x x Ò− − ≤ −
6,
2
1 1 4
3( 98)
x
NN
x
− −
< −
7,
2
2
4( 01)
(1 1)
x
x SPVinh
x
> − −
+ +
8,
2 2
12 12
( 99)
11 2 9
+ < + − −
15,
2 2
( 4) 4 ( 2) 2( 99)x x x x x HVNH− − + + − < −