问题 解答题
选用适当的方法解下列方程:
(1)4(y-1)2=36               
(2)x2-3x-2=0.
(3)(2x+1)2-5(2x+1)+6=0   
(4)3x2-6x-1=0  (配方法)
(5)化简求值:(1+
1
x
)÷
x2-1
x
,x=
2
+1.
答案

(1)4(y-1)2=36,

2(y-1)=±6,

y1=4,y2=-2.

          

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

b2-4ac=(-3)2-4×1×(-2)=17,

x=

17
2×1

x1=

3+
17
2
,x2=
3-
17
2

(3)(2x+1)2-5(2x+1)+6=0,

(2x+1-3)(2x+1-2)=0,

2x+1-3=0,2x+1-2=0,

x1=1,x2=

1
2

   

(4)3x2-6x-1=0,

x2-2x=

1
3

x2-2x+1=

1
3
+1,

(x-1)2=

4
3

x-1=±

2
3
3

x1=

3+2
3
3
,x2=
3-2
3
3

(5)(1+

1
x
)÷
x2-1
x

=

x+1
x
×
x
(x+1)(x-1)

=

1
x-1

当x=

2
+1时,原式=
1
2
+1-1
=
2
2

单项选择题
名词解释