\(y^2+2\left(x^2+1\right)=2\left(x+1\right)\)
\(\Leftrightarrow y^2+2\left(x^2+x+1\right)=2\left(x+1\right)\)
\(\Leftrightarrow y^2+2x^2+2x+2=2x+2\)
\(\Leftrightarrow y^2+2x^2=0\)
Vì \(x^2\ge0;y^2\ge0\)
\(\Rightarrow y^2+2x^2\ge0\)
Mà \(y^2+2x^2=0\)
Nên \(\hept{\begin{cases}y^2=0\\2x^2=0\end{cases}}\)
Hay x = y = 0