x²+x-1=0，所以x²=-(x-1)
x（1-2除以1-x）除以（x+1）-x（x²-1）除以x平方-2x+1
=-x(1+x)除以(1-x)除以(x+1)-x(x+1)(x-1)除以(x-1)²
=x/(x-1)-x(x+1)/(x-1)
=x(1-x-1)/(x-1)
=-x²/(x-1)
=-(x-1)/(x-1)
=-1
x²+x-1=0，x-1=-x²
x（1-2/（1-x））/（x+1）-x（x²-1）/（x²-2x+1）=x[（1-x-2)/(1-x)]/(x+1)-x(x-1)(x+1)/(x-1)²
=x/(x-1)-x(x+1)/（x-1)=x(1-x-1)/(x-1)=-x²/（x-1)=1
x[1-2/(1-x)]/（x+1）]-x（x²-1）/(x²-2x+1)=x(x+1)/(x-1)-x(x²-1)/(x-1)²=x(x+1)/(x-1)-x(x+1)/(x-1)=0.
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