已知x^2+x-1=0,求x(1-2/1-x )÷(x+1)- x(x^2-1)/x^2-2x+1的值
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x²+x-1=0
所以-x²=x-1
原式=x[(1-x-2)/(1-x)÷(x+1)-x(x+1)(x-1)/(x-1)²
=x(-x-1)/(1-x)÷(x+1)-x(x+1)/(x-1)
=x(x+1)/(x-1)÷(x+1)-x(x+1)/(x-1)
=x/(x-1)-(x²+x)/(x-1)
=(x-x²-x)/(x-1)
=-x²/(x-1)
=(x-1)/(x-1)
=1
所以-x²=x-1
原式=x[(1-x-2)/(1-x)÷(x+1)-x(x+1)(x-1)/(x-1)²
=x(-x-1)/(1-x)÷(x+1)-x(x+1)/(x-1)
=x(x+1)/(x-1)÷(x+1)-x(x+1)/(x-1)
=x/(x-1)-(x²+x)/(x-1)
=(x-x²-x)/(x-1)
=-x²/(x-1)
=(x-1)/(x-1)
=1
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