ta có : \(3x^2+3y^2+4xy+2x-2y+2=0\)
\(\Leftrightarrow2x^2+4xy+2y^2+x^2+2x+1+y^2-2y+1=0\)
\(\Leftrightarrow2\left(x+y\right)^2+\left(x+1\right)^2+\left(y-1\right)^2=0\)
ta có : \(2\left(x+y\right)^2+\left(x+1\right)^2+\left(y-1\right)^2\ge0\forall x;y\)
vì vậy : \(2\left(x+y\right)^2+\left(x+1\right)^2+\left(y-1\right)^2=0\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x+1=0\\y-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-y\\x=-1\\y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\) vậy \(x=-1;y=1\)
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