【x-1分之1+1-x^2分之2x】

网友回答

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)
======以下答案可供参考======
供参考答案1:
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)
供参考答案2:
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