问题 解答题
先化简再求值:
(1)(
1
x+2
-
1
x-2
)×(
2
x2+2x
),其中x=6
(2)(
x-1
x
-
x-2
x+1
)×
2x2-x
x2+2x+1
,其中x满足x2-x-1=0.
答案

(1)原式=

x-2-x-2
(x+2)(x-2)
x(x+2)
2

=

-4
(x+2)(x-2)
x(x+2)
2

=-

2x
x-2

当x=6时,原式=-

12
4
=-3;

(2)原式=

x2-1-x2+2x
x(x+1)
(x+1)2
x(2x-1)

=

2x-1
x(x+1)
(x+1)2
x(2x-1)
=
x+1
x2

∵x2-x-1=0,

∴x2=x+1,

则原式=1.

选择题
问答题