题目

已知,y=（x^2-2x+1/x^2-1）/(x^2-x)/(x+1)-(1/x)+1 是说明在右边代数式有意义的天剪下,不论X为何值y值
不变
已知，y=（x^2-2x+1/x^2-1）/(x^2-x)/(x+1)-(1/x)+1 是说明在右边代数式有意义的条件下，不论X为何值y值

提问时间：2021-01-31

答案

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.