\(H=2-5x^2-y^2-4xy+2x\\ =3-5x^2-y^2-4xy+2x-1\\ =3-\left(5x^2+y^2+4xy-2x+1\right)\\ =3-\left[\left(4x^2+4xy+y^2\right)+\left(x^2-2x+1\right)\right]\\ =3-\left[\left(2x+y\right)^2+\left(x-1\right)^2\right]\left(x-1\right)^2\ge0\forall x;\left(2x+y\right)^2\ge0\forall x,y\\ \Rightarrow\left(2x+y\right)^2+\left(x-1\right)^2\ge0\forall x,y\\ \Rightarrow H=3-\left[\left(2x+y\right)^2+\left(x-1\right)^2\right]\le3\forall x,y\\ \text{Dấu }"="\text{ xảy ra khi }\\ \left\{{}\begin{matrix}\left(2x+y\right)^2=0\\\left(x-1\right)^2=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x+y=0\\x-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)Vậy \(Max_A=3\text{ khi }x=1;y=-2\)