问题 解答题
用适当的方法解下列方程(组):
(1)(x-5)2-9=0;
(2)3x2-1=6x
(3)x2+2x-63=0
(4)
2x-1
x
-
3x
2x-1
=2

(5)
x+y=13
(x-1)(y-1)=30

(6)
x2-2xy-3y2=0
x-2y=6
答案

(1)(x-5)2-9=0;

(x-5)2=9,

∴x-5=±3,

∴x1=8,x2=2,

(2)3x2-1=6x,

∴3x2-6x-1=0,

△=b2-4ac=36+12=48,

x=

48
2×3
=
6±4
3
6

∴x1=1+

2
3
3
,x2=1-
2
3
3

(3)x2+2x-63=0,

∴(x-7)(x+9)=0,

∴x1=7,x2=-9,

(4)

2x-1
x
-
3x
2x-1
=2,

先设

2x-1
x
=y,根据题意得:

y-

3
y
=2,

∴y2-2y-3=0,

(y-3)(y+1)=0,

∴y-3=0或y+1=0,

∴y1=3,y2=-1,

2x-1
x
=3,或
2x-1
x
=-1,

∴x1=-1,x2=

1
3

把x1=-1和x2=

1
3
分别代入x-1中,都不等于0,

∴x1=-1,x2=

1
3
是原方程的解;

(5)

x+y=13  ①
(x-1)(y-1)=30 ②

由①得:x=13-y,

∴(13-y-1)(y-1)=30,

∴(12-y)(y-1)=30,

∴y2-13y+42=0,

(y-6)(y-7)=0,

∴y1=6,y2=7,

∴x1=13-6=7,x2=13-7=6,

x1=7
y1=6
x2=6
y2=7

(6)

x2-2xy-3y2=0 ①
x-2y=6   ②

由②得:x=6+2y,

∴(6+2y)2-2(6+2y)y-3y2=0,

y2-4y-12=0,

(y-6)(y+2)=0,

∴y1=6,y2=-2,

∴x1=18,x2=2,

x1=18
y1=6
x2=2
y2=-2

名词解释
判断题