1樓：念含玉
[（x^2-2x）/(x^2-1)] ÷ [x-1-(2x-1)/(x＋1)]
=[（x^2-2x）/(x+1)(x-1)÷ [(x^2-1-2x+1)/(x＋1)]
=[（x^2-2x）/(x+1)(x-1)]×[（x＋1）/(x^2-2x)]
=1/(x-1)
當x1/2時，原式=1/(-1/2)=-2
2樓：金龍
[（x^2-2x）/(x^2-1)] ÷ [x-1-(2x-1)/(x＋1)]
=x（x-2)/(x+1)(x-1)*[(x+1)/(x^2-2x)]
=1/(x-1)
=1/(0.5-1)=-2
3樓：旁悅禕
(x-2)/(x-1)=3
4樓：楊柳
原式=(x^2-2x)/(x^2-1)*(x^2-1-2x+1)/(x+1)=(x^2-2x)/[(x+1)(x-1)]*(x+1)/(x^2-2x)=1/x=2
5樓：匿名使用者
原式=[（x^2-2x）/(x^2-1)] ÷ [(x-1)(x+1)/(x+1)(2x-1)/(x＋1)]
=[（x^2-2x）/(x^2-1)] ÷ [(x^2-2x)/(x＋1)]
=[（x^2-2x）/(x^2-1)] × [(x＋1)/(x^2-2x)]
原式=1/（x-1)
=1/(1/2-1)=-2
6樓：高不成低不就
[（x^2-2x）/(x^2-1)] ÷ [x-1-(2x-1)/(x＋1)]
=x(x-2)/[(x+1)(x-1)]÷[-x/(x+1)]=x(x-2)(x+1)/[(x+1)(x-1)(-x)]=-(x-2)/(x-1)
=-1+1/(x-1)
x=1/2時，原式=-1+1/（1/2-1）=-3
7樓：幻想之魄
(x(x-2)/((x+1)(x-1)))*((x+1)/(x^2-1-2x+1))=1/(x-1)=-2
8樓：匿名使用者
[（x^2-2x）/(x^2-1)] ÷ [x-1-(2x-1)/(x＋1)]
=[（x^2-2x）/(x^2-1)] ÷ [(x-1)(x+1)/(x+1)(2x-1)/(x＋1)]
=[（x^2-2x）/(x^2-1)] ÷ [(x^2-2x)/(x＋1)]
=[（x^2-2x）/(x^2-1)] × [(x＋1)/(x^2-2x)]
=1/（x-1)
=1/(1/2-1)=-2
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