\(3x^2+3y^2+4xy+2x-2y+2=0\)
\(\Leftrightarrow\left(2x^2+2y^2+4xy\right)+\left(x^2+2x+1\right)+\left(y^2-2y+1\right)=0\)
\(\Leftrightarrow2\left(x^2+y^2+2xy\right)+\left(x+1\right)^2+\left(y-1\right)^2=0\)
\(\Leftrightarrow2\left(x+y\right)^2+\left(x+1\right)^2+\left(y-1\right)^2=0\)
Vì \(\left(x+y\right)^2\ge0\); \(\left(x+1\right)^2\ge0\); \(\left(y-1\right)^2\ge0\)\(\forall x,y\)
\(\Rightarrow2\left(x+y\right)^2+\left(x+1\right)^2+\left(y-1\right)^2\ge0\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x+y=0\\x+1=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-y\\x=-1\\y=1\end{cases}}\)
Vậy \(x=-1\)và \(y=1\)