已知x=根号2+1，求（x+1/x的平方-x-x/x的平方-2x+1）/1/x,详细过程

已知x=根号2+1，求（x+1/x的平方-x-x/x的平方-2x+1）/1/x,详细过程

解：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

(x^2-√2x+1)(x^2+√2x+1)-(x^2-x+1)(x^2+x+1) =[(x^2+1)^2-(√2x)^2]-[(x^2+1)^2-x^2] =(x^2+1)^2-(√2x)^2-(x^2+1)^2+x^2 =x^2-(√2x)^2 =x^2-2x^2 =-x^2<=0 所以(x^2-√2x+1)(x^2+√2x+1)<=(x^2-x+1)(x^2+x+1)

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