\(A=-x^2-4y^2+2x-12y-10\)
\(A=-\left(x^2-2x+1\right)-\left(4y^2-12y+9\right)\)
\(A=-\left(x-1\right)^2-\left(2y+3\right)^2\)
Vậy\(A_{max}=0\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-\dfrac{3}{2}\end{matrix}\right.\)
Ta có: \(-x^2+2x-4y^2-12y-10\)
\(=-\left(x^2-2x+1+4y^2+12y+9\right)\)
\(=-\left(x-1\right)^2-\left(2y+3\right)^2\le0\forall x,y\)
Dấu '=' xảy ra khi \(\left(x,y\right)=\left(1;-\dfrac{3}{2}\right)\)