y=[(x²-2x+1)/(x²-1)]/[(x²-x)/(x+1)] -(1/x) +1
=(x²-2x+1)(x+1)/[(x²-1)(x²-x)] +(x-1)/x
=(x-1)²(x+1)/[x(x-1)(x-1)(x+1)] +(x-1)/x
=1/x +(x-1)/x
=(x-1+1)/x
=x/x
=1
无论x取何使得代数式有意义的实数值,y的值恒为1,为定值.