Question

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

已知x²+x-1=0，求x(1-2/1-x)除号（x+1）-（x²-1/x²-2x+1）的值

Answer 1

x²+x-1=0
则-x²=x-1

原式=[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)
=(x-1)/(x-1)
=1

追问

已知x²+x-1=0，求x(1-2/1-x)除号（x+1）-（x²-1/x²-2x+1）的值
处理提问

追答

x[1-(2/1-x)]÷(x+1)-(x^2-1)/（x^2-2x+1）
=x[(1-x-2)/(1-x)]÷(x+1)-(x^2-1)/(x^2-2x+1)
=x/(x-1)-(x+1)(x-1)/(x-1)²
=x/(x-1)-(x+1)/(x-1)
=-1/(x-1)
∵x^2+x-1=0
x^2+x+1/4-5/4=0
(x+1/2)²=5/4
x=2分之（±根号5-1）
原式=-2分之（根号5+3）或者-2分之（根号5-3）