将m(a-x)(a-x)(x-b)-nm(x-a)(b-x)(b-x)分解因式,可提取公因式什么?
问题描述：
将m(a-x)(a-x)(x-b)-nm(x-a)(b-x)(b-x)分解因式,可提取公因式什么?
答
m(a-x)(b-x)
答
当然可以，
m(a-x)(a-x)(x-b)-nm(x-a)(b-x)(b-x)
=m(a-x)(a-x)(x-b)+nm(a-x)(x-b)(x-b)
=m(a-x)(x-b)[a-x+n(x-b)]
=m(a-x)(x-b)[(n-1)x-nb+a]
答
=m(a-x)(a-x)(x-b)-mn(a-x)(x-b)(b-x)
=m(a-x)(x-b）[a-x-n(b-x)]
=m(a-x)(x-b）（a-x-bn-nx)
答
= m(a-x)(a-x)(x-b)-nm(a-x)(x-b)(b-x)
= m(a-x)(x-b)[(a-x)-n(b-x)]
答
提m(x-a)(x-b)结果为m(x-a)(x-b)[(1-n)x-a+nb].