问题描述: 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. 1个回答 分类:数学 2014-11-06 问题解答: 我来补答 y=x^2-2x+1/x^2-1除x^2-x/x+1*5xy=(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. 展开全文阅读