m^2(2次方）－n^2+3m-3n＝0,求（m^3n-2m^2n^2=mn^3)/(2mn)+2mn的值
问题描述：
m^2(2次方）－n^2+3m-3n＝0,求（m^3n-2m^2n^2=mn^3)/(2mn)+2mn的值
答
m^2-n^2+3m-3n=0
(m-n)(m+n+3)=0
得
m=n
m+n=-3
得
m=n=-3/2
（m^3n-2m^2n^2+mn^3)/(2mn)+2mn
=[mn(m^2-2mn+n^2)]/2mn+2mn
=(m-n)^2/2+2mn
=2*(-3/2)^2
=9/2
答
(m^3n-2m^2n^2+mn^3)/(2mn)+2mn=(m^3n-2m^2n^2+mn^3+4m^2n^2)/(2mn)
=(m^3n+2m^2n^2+mn^3)/(2mn)=(m+n)^2/2
m^2－n^2+3m-3n=(m-n)(m+n+3)=0,有m=n或m+n=-3
则原式=9/2
答
m^2-n^2+3m-3n＝0(m+n)(m-n)-3(m-n)=0(m-n)(m+n-3)=0m-n=0或m+n-3=0（m^3n-2m^2n^2+mn^3)/(2mn)+2mn=（m^3n-2m^2n^2+mn^3+4m^2n^2)/(2mn)=（m^3n+2m^2n^2+mn^3)/(2mn)=mn(m+n)^2/2mn=(m+n)^2/2若m-n=0,则m=n原式=2m^...
答
题出错啦