\(x^2+2y^2-2xy+4y+4=0\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2+4y+4\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y+2\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x-y=0\\y+2=0\end{cases}}\Leftrightarrow x=y=-2\)
Vậy \(x+y=-2-2=-4\)