\(3x^2+y^2+2x-2y=1\)
\(\Leftrightarrow\left(2x^2+2xy\right)+x^2-2xy+y^2+2\left(x-y\right)+1=2\)
\(\Leftrightarrow A+\left(x-y+1\right)^2=2\)
\(A>0=\left(x-y+1\right)^2< 2\)
A nguyên \(\orbr{\begin{cases}\left(x-y+1\right)^2=0\\\left(x-y+1\right)^2=1\end{cases}\Rightarrow\orbr{\begin{cases}A=2\\A=1\end{cases}}}\)