已知x²+x-1=0,求x（1-2除以1-x）除以（x+1）-x（x²-1）除以x平方-2x+1的值

已知x²+x-1=0,求x（1-2除以1-x）除以（x+1）-x（x²-1）除以x平方-2x+1的值
已知x²+x-1=0,求x（1-2除以1-x）除以（x+1）-x（x²-1）除以x平方-2x+1的值

已知x²+x-1=0,求x（1-2除以1-x）除以（x+1）-x（x²-1）除以x平方-2x+1的值
x²+x-1=0,则有：-x²=x-1
x[1-2/(1-x)]÷(x+1）-x(x²-1)/(x²-2x+1)
=x(x+1)/(x-1)(x+1) - x(x-1)(x+1)/(x-1)²
=x/(x-1) - x(x+1)/(x-1)
=-x²/(x-1)=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-2)/(1-x)]/(x+1)-x(x+1)(x-1)/(x-1)²
=x/(x-1)-x(x+1)/(x-1)
=(x-x²+x)/(x-1)
=x²/(x-1)
=(1-x)/(x-1)
=-1

-1或-（2+根号5）