把分式化简成整式！ 求过程

Question

x-1分之1+1-x^2分之2x

Answer 1

解：
原式=
(1/(x-1))+(2x/(1-x^2))

(1/(x-1))在分式上下同乘(x+1),即用分式的基本性质
得:
((x+1)/(x^2-1))+(2x/(1-x^2))

(2x/(1-x^2))在分式上下同乘(-1),即用分式的基本性质
得:
(x+1-2x)/(x^2-1)
=(1-x)/(x+1)(x-1) (逆用平方差公式)
= -(x-1)/(x+1)(x-1)

在分式上下同除(x-1),即用分式的基本性质
得:
=(-1)/(x+1)
答:为(-1)/(x+1)本回答被提问者采纳

Answer 2

原式=（1/（x+1))-(2x/(x-1)(x+1))
=(x+1-2x)/(x-1)(x+1)
=(1-x)/(x-1)(x+1)
=1/(-x-1)

Answer 3

2X/[1/(X-1)+1/(1-X^2)]=2X/[1/(X-1)+1/(1+X)(1-X)]=2X/[(1-1-X)/(X-1)]=2X/[(-X)/(X-1)]=2X*[(1-X)/X]=2-2X