1/(x＋1)＋(x²－2x＋1)/(x²－1)除以(x－1)/(x＋1)
=1/(x＋1)＋(x－1)²/[(x－1)(x＋1)]除以(x－1)/(x＋1)
=1/(x＋1)＋(x－1)/(x＋1)除以(x－1)/(x＋1)
=[x/(x＋1)]×[(x＋1)/(x－1)]
=x/(x－1)
当x=2时，原式=2
(x+1分之1 + x方-1分之x方-2x+1)÷x+1分之x-1
=[1/(x+1)+(x-1)²/(x-1)(x+1)]÷(x-1)/(x+1)
=[x/(x+1)]×(x+1)/(x-1)
=x/(x-1) 其中x=2
=2/(2-1)
=2
[1/(x＋1)＋(x²－2x＋1)/(x²－1)] / [(x－1)/(x＋1)]
= [1/(x＋1)＋(x－1)²/[(x－1)(x＋1)] ]/ [ (x－1)/(x＋1)]
=[x/(x＋1)]×[(x＋1)/(x－1)]
=x/(x－1)
将x=2代入式中得2/（2-1）=2
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