题目

若多项式2x^2+mx-y+6与2nx^2-3x+5y-1的差的值与x所取的值无关,试求多项式1/3m^2-2n^2-(1/4m^2-3n^2)的值

提问时间：2020-12-21

答案

2x^2+mx-y+6-(2nx^2-3x+5y-1)
=(2-2n)x^2+(m+3)x-6y+7
因为与x所取的值无关,所以2-2n=0,m+3=0.或m=-3,n=1.
1/3m^2-2n^2-(1/4m^2-3n^2)
=(1/3-1/4)(-3)^2-(2-3)1^2
=3/4+1
=7/4