化简(2x^2+2x/x^2-1-x^2-x/x^2-2x+1)÷x/x+1,当x=1+√2,原代数式的值能等于-1吗

(2x^2+2x/x^2-1-x^2-x/x^2-2x+1)÷x/x+1
当x=1+√2 原式=[2x(x+1)/(X+1)(X-1)-(x-1)x/（x-1)²]乘以（x+1)/x
={[2X/(X-1)]- [x/（x-1）]}×（x+1)/x= x/(x-1) × （x+1)/x =(x+1)/(x-1)
带入1+√2得出原式=√2+1