求证:多项式(x-1)(x-2)(x-3)(x-4)-(x^2-5x)^2-10(x^2-5x)的值与x的取值无关

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/08 19:30:13
求证:多项式(x-1)(x-2)(x-3)(x-4)-(x^2-5x)^2-10(x^2-5x)的值与x的取值无关

求证:多项式(x-1)(x-2)(x-3)(x-4)-(x^2-5x)^2-10(x^2-5x)的值与x的取值无关
求证:多项式(x-1)(x-2)(x-3)(x-4)-(x^2-5x)^2-10(x^2-5x)的值与x的取值无关

求证:多项式(x-1)(x-2)(x-3)(x-4)-(x^2-5x)^2-10(x^2-5x)的值与x的取值无关
(x-1)(x-2)(x-3)(x-4)-(x^2-5x)^2-10(x^2-5x)
=[(x^2-5x)+4][(x^2-5x)+6]-(x^2-5x)^2-10(x^2-5x)
=(x^2-5x)^2+10(x^2-5x)+24-(x^2-5x)^2-10(x^2-5x)
=24
所以多项式(x-1)(x-2)(x-3)(x-4)-(x^2-5x)^2-10(x^2-5x)的值与x的取值无关

就是把原式直接乘开化简成为一个常数就OK了。就说明多项式的值与x的取值无关。
分解因式是行不通的(试过)只有这种方法了。需要我做一遍吗?

楼主你好
可以把多项式展开
(x-1)(x-2)(x-3)(x-4)-(x^2-5x)^2-10(x^2-5x)=(x^2-3x+2)(x^2-7x+12)-(x^4-10x^3+25x^2)-(10x^2-50x)=x^4-10x^3+35x^2-50x+24-x^4+10x^3-25x^2-10x^2+50x=24
所以(x-1)(x-2)(x-3)(x-4)-(x^2-5x)^2-10(x^2-5x)的值与x的取值无关
希望你满意