\(A=-10x^2-y^2+6xy-4x+20\)
\(=-9x^2+6xy-y^2-x^2-4x-4+24\)
\(=-\left(3x-y\right)^2-\left(x+2\right)^2+24\le24\forall x,y\)
Dấu '=' xảy ra khi \(\begin{cases}3x-y=0\\ x+2=0\end{cases}\Rightarrow\begin{cases}x=-2\\ y=3x=3\cdot\left(-2\right)=-6\end{cases}\)