Ta có : \(x^2+2y^2+2xy+2x+4y+2=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(2x+2y\right)+1+\left(y^2+2y+1\right)=0\)
\(\Leftrightarrow\left(x+y\right)^2+2\left(x+y\right)+1+\left(y+1\right)^2=0\)
\(\Leftrightarrow\left(x+y+1\right)^2+\left(y+1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y+1\right)^2=0\\\left(y+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y+1=0\\y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=-1\end{matrix}\right.\)
Vậy \(x=0\) và \(y=-1\)
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