已知x=根号2+1,求（x+1/x的平方-x-x/x的平方-2x+1）/1/x,
问题描述：
已知x=根号2+1,求（x+1/x的平方-x-x/x的平方-2x+1）/1/x,
答
x=√2+1
原式=[(x+1)/(x²-x)-x/(x²-2x+1)]÷(1/x)
={(x+1)/[x(x-1)]-x/(x-1)²}×x
={(x+1)(x-1)/[x(x-1)²]-x²/[x(x-1)²]}×x
={(x²-1-x²)/[x(x-1)²]}×x
=-1/(x-1)²
=-1/(√2+1-1)²
=-1/(√2²)
=-1/2