\(B=-4x^2-5y^2+8xy+10y+12\)
\(=-4x^2+8xy-4y^2-y^2+10y-25+37\)
\(=-\left(2x-2y\right)^2-\left(y-5\right)^2+37< =37\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}2x-2y=0\\y-5=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=5\\x=y=5\end{matrix}\right.\)