先化简,再求值:(x−1x−x−2x+1)÷2x2−xx2+2x+1,其中x满足x2-x-1=0.

问题描述:

先化简,再求值:(

x−1
x
x−2
x+1
2x2−x
x2+2x+1
,其中x满足x2-x-1=0.

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

∵x2-x-1=0,
∴x2=x+1,
将x2=x+1代入化简后的式子得:
x+1
x2
=
x+1
x+1
=1.