bai tap he phuong trinh
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Bµi tËp hÖ ph-¬ng tr×nh
Gi¶i c¸c hÖ ph-¬ng tr×nh sau :
x + xy + y = −1
1, 2
( MTCN − 99)
2
x y + y x = −6
x 2 y + y 2 x = 30
3, 3
( BK − 93)
3
x + y = 35
x 2 + y 2 + xy = 7
5, 4
(SP1 − 2000)
4
2 2
x + y + x y = 21
x 2 + y 2 = 5
(NT − 98)
2, 4
2 2
4
x − x y + y = 13
x 3 + y 3 = 1
4, 5
( AN − 97)
5
2
2
x + y = x + y
x + y + xy = 11
6, 2
(QG − 2000)
2
x + y + 3( x + y ) = 28
x
7
y
+
=
+1
x
xy
7, y
( HH − 99)
x xy + y xy = 78
1
( x + y )(1 + xy ) = 5
(NT − 99)
8,
( x 2 + y 2 )(1 + 1 ) = 49
x2 y2
1 1
x + y + x + y = 4
( AN − 99)
9,
1
1
2
2
x + y + +
=4
x 2 y2
x ( x + 2)(2 x + y ) = 9
10, 2
( AN − 2001)
x + 4x + y = 6
x 2 + x + y + 1 + x + y 2 + x + y + 1 + y = 18
( AN − 99)
11,
x 2 + x + y + 1 − x + y 2 + x + y + 1 − y = 2
y + xy 2 = 6 x 2
x (3 x + 2 y )( x + 1) = 12
( SP1 − 2000)
12, 2
13,
( BCVT − 97)
2 2
2
1 + x y = 5 x
x + 2y + 4 x − 8 = 0
2 x 2 − 3 x = y 2 − 2
x + y = 4
(QG − 2000)
14, 2
( HVQHQT − 2001) 15, 2
2
3
3
2
2 y − 3y = x − 2
( x + y )( x + y ) = 280
1 3
2x + =
2
x = 3 x − y
y x
( MTCN − 98)
(QG − 99)
16, 2
17,
1
3
y = 3 y − x
2 y + =
x y
x = 3 x + 8y
18, 3
(QG − 98)
y = 3y + 8 x
3
x + 5 + y − 2 = 7
(NN1 − 2000)
20,
y + 5 + x − 2 = 7
3
2 x + y = x 2
( TL − 2001)
19,
3
2 y + x =
y2
y2 + 2
3 y = x 2
21,
( KhèiB − 2003)
2
3 x = x + 2
y2
GV:NGUYỄN MINH NHIÊN-TRƯỜNG THPT QUẾ VÕ 1
3 x 2 2 xy = 16
22, 2
( HH TPHCM )
2
x 3 xy 2 x = 8
2
2
x 2 xy + 3 y = 9
( HVNH TPHCM )
24, 2
2
2 x 13 xy + 15 y = 0
1 + x 3 y 3 = 19 x 3
23,
(TM 2001)
2
2
y + xy = 6 x
2
2
2 y( x y ) = 3 x
( M Đ C 97)
25, 2
2
x ( x + y ) = 10 y
Bài tập ph-ơng trình -bất ph-ơng trình vô tỉ
Giải các ph-ơng trình sau:
1, x + 3 + 6 x = 3
2, x + 9 = 5 2 x + 4
3, x + 4 1 x = 1 2 x
4, ( x 3) 10 x 2 = x 2 x 12
5, 3 x + 4 3 x 3 = 1
6, 3 2 x 1 + 3 x 1 = 3 3 x + 1
7, 2 x + 2 + x + 1 x + 1 = 4(khốiD 2005)
8, x + 2 x 1 x 2 x 1 = 2( BCVT 2000)
9, 3(2 + x 2 ) = 2 x + x + 6( HVKTQS 01)
10, 2 x 2 + 8 x + 6 + x 2 1 = 2 x + 2( BK 2000)
11,
5
5
x2 + 1 x2 +
x 2 1 x 2 = x + 1( PCCC 2001)
4
4
12, x ( x 1) + x ( x + 2) = 2 x 2 (SP2 2000 A)
13, 2 x 2 + 8 x + 6 + x 2 1 = 2 x + 2( HVKTQS 99)
Tìm m để ph-ơng trình :
14, x 2 + mx + 2 = 2 x + 1( KhốiB 2006)
có 2 nghiệm phân biệt
15, 2 x 2 + mx = 3 x ( SPKT TPHCM )
có nghiệm
16, 2 x 2 + mx 3 = x m(GT 98)
Giải các ph-ơng trình sau :
có nghiệm
17, x 2 + x 2 + 11 = 31
18, ( x + 5)(2 x ) = 3 x 2 + 3 x
19, x 2 3 x + 3 + x 2 3 x + 6 = 3( TM 98)
20, 2 x 2 + 5 x 1 = 7 x 3 1
21, x 2 + 2 x + 4 = 3 x 3 + 4 x
22, 3 x + x 2 2 + x x 2 = 1(NT 99)
23, x + 1 + 4 x + ( x + 1)(4 x )(NN 20001)
24, x + 4 x 2 = 2 + 3 x 4 x 2 ( M Đ C 2001)
25, x 2 + 4 x = x 2 6 x + 11
26, 2 x 3 + 5 2 x + 4 x x 2 6 = 0( GTVT TPHCM 01)
27, 3 x 2 + x 1 = 4 x 9 + 2 3 x 2 5 x + 2( HVKTQS 97)
x2 + 7x + 4
28,
= 4 x ( DL Đông Đô2000)
x+2
29, 3
2x
1 1
+3 +
= 2( GT 95)
x +1
2 2x
GV:NGUYN MINH NHIấN-TRNG THPT QU Vế 1
30, x +
x
x −1
2
=2 2
31, 1 + 1 − x 2 = x (1 + 2 1 − x 2 )
32, (4 x − 1) x 2 + 1 = 2 x 2 + 2 x + 1(§ Ò 78)
33, x 2 + 3 x + 1 = ( x + 3) x 2 + 1( GT − 01)
34, 2(1 − x ) x 2 + 2 x − 1 = x 2 − 2 x − 1
35, x 2 + x + 1 = 1( XD − 98)
36, 3 2 − x = 1 − x − 1( TCKT − 2000)
37, 3 x + 7 − x = 1( LuËt − 96)
7− x − 3 x −5
= 6 − x (C § − KiÓmS ¸ t )
3
7− x + 3 x −5
Gi¶i c¸c bÊt ph-¬ng tr×nh sau :
1, ( x − 1)(4 − x ) > x − 2( M § C − 2000)
3
38,
3, x + 3 ≥ 2 x − 8 + 7 − x ( AN − 97)
5, ( x − 3) x 2 − 4 ≤ x 2 − 9(§ Ò 11)
7,
x2
> x − 4(SPVinh − 01)
(1 + x + 1)2
39, x 3 + 1 = 2 3 2 x − 1
2, x + 1 > 3 − x + 4( BK − 99)
4, x + 2 − 3 − x < 5 − 2 x ( TL − 2000)
1 − 1 − 4x2
< 3(NN − 98)
x
12 + x − x 2
12 + x − x 2
≥
( HuÕ − 99)
8,
x − 11
2x − 9
6,
9, x 2 + 3 x + 2 + x 2 + 6 x + 5 ≤ 2 x 2 + 9 x + 7( BK − 2000)
10, x 2 − 4 x + 3 − 2 x 2 − 3 x + 1 ≥ x − 1( KT − 2001)
11, 5 x 2 + 10 x + 1 ≥ 7 − x 2 − 2 x (§ Ò 135)
12, −4 (4 − x )(2 + x ) ≤ x 2 − 2 x − 12(§ Ò 149)
13, ( x 3 + 1) + ( x 2 + 1) + 3 x x + 1 > 0( XD − 99)
3
1
< 2x +
− 7( Th¸iNguy ª n − 2000)
14, 3 x +
2x
2 x
15, x ( x − 4) − x 2 + 4 x + ( x − 2)2 < 2( HVNH − 99)
GV:NGUYỄN MINH NHIÊN-TRƯỜNG THPT QUẾ VÕ 1
