当x取什么值时,能使分式(x^4+x^3-2)/(x^3-x^2+x-1)*(x^4-1)/(x^2+2x^2+2x+2)÷(x^3-x-x^2+1)/(-2)

来源:百度知道 编辑:UC知道 时间:2024/06/01 05:22:14
当x取什么值时,能使分式(x^4+x^3-2)/(x^3-x^2+x-1)*(x^4-1)/(x^2+2x^2+2x+2)÷(x^3-x-x^2+1)/(-2)的值为正整数

(x^4+x^3-2)/(x^3-x^2+x-1)*(x^4-1)/(x^3+2x^2+2x+2)÷(x^3-x-x^2+1)/(-2)
=[(x-1)(x^3+2x^2+2x+2)/(x-1)(x^2+1)]*[(x^2+1)(x+1)(x-1)/(x^3+2x^2+2x+2)]*[-2/(x-1)^2(x+1)]
=-2(x-1)(x^3+2x^2+2x+2)(x^2+1)(x+1)(x-1)/[(x-1)(x^2+1)(x^3+2x^2+2x+2)(x-1)^2(x+1)]
=-2/(x-1)是正整数
所以x-1=-1,x-1=-2
x=0,x=-1

没有下文啦?

题目不完整啊