已知y=x^2-2x+1/x^2-1除x^2-x/x+1*5x试说明在代数式有意义的条件下,不论x取何值,y的值都不变._数学解答

已知y=x^2-2x+1/x^2-1除x^2-x/x+1*5x试说明在代数式有意义的条件下,不论x取何值,y的值都不变.

人气：282 ℃　时间：2019-08-20 23:09:01

解答

y=x^2-2x+1/x^2-1除x^2-x/x+1*5x
y=(x^2-x/5x(x+1))÷(x^2-2x+1/x^2-1)
y=(x(x-1)/5x(x+1))÷((x-1)^2/(x+1)(x-1))
y=((x-1)/5(x+1))÷((x-1)/(x+1))
y=((x-1)/5(x+1))*((x+1)/(x-1))
y=1/5
所以不论x取何值,y的值都不变.