计算(2m-n)^2*(n^2+4m^2)^2*(2m+n)*2要解题过程
问题描述：
计算(2m-n)^2*(n^2+4m^2)^2*(2m+n)*2要解题过程
答
(2m-n)^2*(n^2+4m^2)^2*(2m+n)*2
题目不对吧应该是(2m-n)^2*(n^2+4m^2)^2*(2m+n)^2
=(2m-n)^2*(2m+n)^2*(n^2+4m^2)^2
=[(2m-n)*(2m+n)]^2*(n^2+4m^2)^2
=(4m^2-n^2)^2*(4m^2+n^2)^2
=[(4m^2-n^2)*(4m^2+n^2)]^2
=(16m^4-n^4)^2
=256m^8-32m^4n^4+n^8
答
(2m-n)^2*(n^2+4m^2)^2*(2m+n)^2
=[(2m-n)*(2m+n)]^2*(n^2+4m^2)^2
=(4m^2-n^2)^2*(n^2+4m^2)^2
=[(4m^2-n^2)*(n^2+4m^2)]^2
=(16m^4-n^4)^2
=256m^8-32(mn)^4-n^8.
答
(2m-n)^2*(n^2+4m^2)^2*(2m+n)*2=(2m-n)^2*(2m+n)*2*(n^2+4m^2)^2=[(2m-n)*(2m+n)]*2*(n^2+4m^2)^2=(4m^2-n^2)*2*(4m^2+n^2)^2=[(4m^2-n^2)*(4m^2+n^2)]^2=(16m^4-n^4)^2=256m^8-32m^4n^4+n^8
答
如果括号里是4*m^2的话，那么第一项和第三项合并，原式=(n^2+4m^2)^2*(4m^2-n^2)^2=(16m^4-n^4)^2=256m^8-32m^4n^4+n^8