题目内容

已知x2+x-1=0,求x(1-
2
1-x
)÷(x+1)-
(x2-1)
x2-2x+1
的值.
考点:分式的化简求值
专题:
分析:先化简所给的代数式,然后解所给的方程,灵活运用解求值即可.
解答:解:x(1-
2
1-x
)÷(x+1)-
(x2-1)
x2-2x+1

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

=-x×
1+x
1-x
×
1
1+x
-
x+1
x-1

=-
x
1-x
-
x+1
x-1

=
x
x-1
-
x+1
x-1

=-
1
x-1

=
1
1-x

解方程x2+x-1=0,得
x1=
-1+
5
2
x2=
-1-
5
2

x2=
5
2
1
x2
=
5
2

∵x2+x-1=0,
∴1-x=x2
∴原式=
1
1-x

=
1
x2

=
3+
5
2
3-
5
2
点评:考查了分式的化简与求值问题:解题的关键是正确化简所给的分式,灵活变形求值.
练习册系列答案
相关题目

违法和不良信息举报电话:027-86699610 举报邮箱:58377363@163.com

精英家教网