Đáp án:
$\begin{array}{l}
{x^2} + 2{y^2} + 2xy + 2x + 6y + 1 = 0\\
\Rightarrow {x^2} + {y^2} + 1 + 2x + 2y + 2xy\\
+ {y^2} + 4y + 4 = 4\\
\Rightarrow {\left( {x + y + 1} \right)^2} + {\left( {y + 2} \right)^2} = 4 = 0 + 4\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
{\left( {x + y + 1} \right)^2} = 4\\
{\left( {y + 2} \right)^2} = 0
\end{array} \right.\\
\left\{ \begin{array}{l}
{\left( {x + y + 1} \right)^2} = 0\\
{\left( {y + 2} \right)^2} = 4
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + y + 1 = 2/x + y + 1 = - 2\\
y = - 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x + y + 1 = 0\\
\left[ \begin{array}{l}
y = 0\\
y = - 4
\end{array} \right.
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 3;y = - 2\\
x = - 1;y = - 2\\
x = - 1;y = 0\\
x = 3;y = - 4
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( {3; - 2} \right);\left( { - 1; - 2} \right);\left( { - 1;0} \right);\left( {3; - 4} \right)} \right\}
\end{array}$
