> 数学 >
先化简,再求值:
(1)(1−
1
a−1
a2−4a+4
a2−a
,其中a=-1;
(2)(
x−1
x
-
x−2
x+1
)÷
2x2−x
x2+2x+1
,其中x满足x2-x-1=0.
人气:472 ℃ 时间:2020-05-12 12:15:04
解答
(1)原式=
a−2
a−1
×
a(a−1)
(a−2)2

=
a
a−2

当a=1时,原式=
1
1−2
=-1;
(2)原式=
(x−1)(x+1)−x(x−2)
x(x+1)
×
(x+1)2
x(2x−1)

=
2x−1
x(x+1)
×
(x+1)2
x(2x−1)

=
x+1
x2

∵x满足x2-x-1=0,
∴x2=x+1,
∴原式=
x+1
x+1
=1.
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