Bµi tËp hÖ ph¬ng tr×nh
Gi¶i c¸c hÖ ph¬ng tr×nh sau :
−
1
6
x xy y
MTCN
+ =
5
( 98) 13
x y
NT
x x y y
3, + =
−
3 3
30 ( 93) 35
x y y x
BK
+ =
1
( 97)
x y
AN
x y x y
−
7 ( 1 2000) 21
x y xy
SP
−
11
( 2000) 3( ) 28
x y xy
QG
x y x y
7,
−
7 1 ( 99) 78
x y
y x xy HH
x xy y xy
8,
1 ( )(1 ) 5
( 99) 1
x y
xy
NT
x y
x y
9,
+ + + =
+ + + =
1 1
4 ( 99)
1 1
4
x y
x y
AN
x y
x y
10, + + =
−
2
( 2)(2 ) 9
( 2001)
x x x y
AN
x x y
−
( 99)
x x y x y x y y
AN
x x y x y x y y
−
2
(3 2 )( 1) 12
2 4 8 0
x x y x
BCVT
6 ( 1 2000)
y xy x
SP
x y x
14, + =
−
4
x y
HVQHQT
( 2000)
x x y
QG
y y x
16, = −
−
2
2
3
3
x x y
MTCN
+ =
+ =
1 3 2
( 99)
1 3 2
x
y x QG y
x y
18, = +
−
3
3
3 8
( 98)
3 8
x x y
QG
+ =
+ =
2
2
3 2
( 2001) 3
2
x y
x TL
y x
y
20, + + − =
−
( 1 2000)
NN
=
2
2
2
2
2 3
2 3
y y
x KhèiB x
x y
22, − =
−
2
x xy
HH TPHCM
( 2001) 6
x y x
TM
y xy x
24, − + =
−
x xy y
HVNH TPHCM
( § 97) ( ) 10
y x y x
M C
x x y y
Bµi tËp ph¬ng tr×nh -bÊt ph¬ng tr×nh v« tØ
Gi¶i c¸c ph¬ng tr×nh sau:
Trang 21, x+ +3 6− =x 3 2, x+ = −9 5 2x+4
(x−3) 10−x =x − −x 12
5,3 x+ −4 3 x− =3 1 6,32x− +1 3 x− =1 33x+1
7, 2 x+ +2 x+ −1 x+ =1 4(khốiD−2005) 8,
x+ x− − x− x− = BCVT−
9, 3(2+ x− =2) 2x+ x+6(HVKTQS−01)
2x +8x+ +6 x − =1 2x+2(BK−2000)
4− +x −x + 4− −x −x = +x PCCC−
( 1) ( 2) 2 ( 2 2000 )
x x− + x x+ = x SP − A
13, 2x2+8x+ +6 x2 − =1 2x+2(HVKTQS−99)
Tìm m để phơng trình :
14, x2+mx+ =2 2x+1(KhốiB−2006) có 2 nghiệm phân biệt
15, 2
2x +mx = −3 x SPKT TPHCM( − ) có nghiệm
16, 2x2+mx− = −3 x m GT( −98) có nghiệm
Giải các phơng trình sau :
17, 2 2
11 31
x + x + = 18,(x+5)(2− =x) 3 x2+3x
19, x2−3x+ +3 x2−3x+ =6 3(TM−98) 20,2x2+5x− =1 7 x3−1
21,x2 +2x+ =4 3 x3+4x 22,
3− +x x − 2+ −x x =1(NT−99)
23, x+ +1 4− +x (x+1)(4−x NN)( −20001)
24,x+ 4−x2 = +2 3x 4−x M C2( Đ −2001)
25, x− +2 4− =x x2−6x+11
26, 2x− +3 5 2− x +4x x− − =2 6 0(GTVT TPHCM− −01)
27, 3x− +2 x− =1 4x− +9 2 3x2−5x+2(HVKTQS−97)
28, 2 7 4 4 ( Đô Đô 2000)
2
x x x DL ng
x
2( 95)
x + + x = − +
1
x
x
x
− 31, 1+ 1−x2 =x(1 2 1+ −x2)
32,(4x−1) x2+ =1 2x2 +2x+1(Đ 78)ề 33,x2+3x+ = +1 (x 3) x2+1(GT−01)
34,2(1−x x) 2+2x− =1 x2−2x−1 35,x2+ x+ =1 1(XD−98)
36,32− = −x 1 x−1(TCKT−2000) 37,3 x+ −7 x =1(Luật−96)
38,3 3
x x
x C KiểmS t
x x
3 1 2 23 1
x + = x−
Giải các bất phơng trình sau :
1, (x−1)(4−x)> −x 2( ĐM C−2000) 2, x+ > −1 3 x+4(BK−99)
Trang 33, x+ ≥3 2x− +8 7−x AN( −97) 4, x+ −2 3− <x 5 2 (− x TL−2000)
(x−3) x − ≤4 x −9(§ 11)Ò 6,1 1 4x2 3( 98)
NN x
7,
2
x x SPVinh
+ + 8,
x x x x
HuÕ
x + x+ + x + x+ ≤ x + x+ BK−
10, x2−4x+ −3 2x2−3x+ ≥ −1 x 1(KT−2001)
5x +10x+ ≥ − −1 7 x 2 (§ 135)x Ò
12, −4 (4−x)(2+x)≤x2−2x−12(§ 149)Ò
(x + +1) (x + +1) 3x x+ >1 0(XD−99)
14, 3 3 2 1 7( ¸ ª 2000)
2 2
x x Th iNguy n
x x
x x− − +x x + −x < HVNH−