已知y=x^2-2x+1/x^2-1除以x^2-x/x+1-1/x+1,已知y=x^2-2x+1/x^2-1除以x^2-x/x+1_数学解答

已知y=x^2-2x+1/x^2-1除以x^2-x/x+1-1/x+1,
已知y=x^2-2x+1/x^2-1除以x^2-x/x+1-1/x+1,试说明在右边代数式有意义的条件下,不论x为何值,y值都不变

人气：322 ℃　时间：2020-02-06 02:29:51

解答

y=[(x-1)^2/(x+1)(x-1)]÷[x(x-1)/(x+1)]-1/x+1
=[(x-1)/(x+1)]×[(x+1)/x(x-1)]-1/x+1
=1/x-1/x+1
=1
所以不论x为何值,y值都不变