\(x-y=2\Rightarrow x=y+2\)
\(\Rightarrow A=2\left(y+2\right)^2+y^2=3y^2+8y+8\)
\(A=3\left(y^2+\frac{8}{3}y+\frac{16}{9}\right)+\frac{8}{3}=3\left(y+\frac{4}{3}\right)^2+\frac{8}{3}\ge\frac{8}{3}\)
\(\Rightarrow A_{min}=\frac{8}{3}\) khi \(\left\{{}\begin{matrix}y=-\frac{4}{3}\\x=y+2=\frac{2}{3}\end{matrix}\right.\)