Đáp án:
$\begin{array}{l}
{x^2} + 2{y^2} - 2xy - 2x - 3 = 0\\
\Rightarrow 2{x^2} + 4{y^2} - 4xy - 4x - 6 = 0\\
\Rightarrow \left( {{x^2} - 4xy + 4{y^2}} \right) + \left( {{x^2} - 4x + 4} \right) = 10\\
\Rightarrow {\left( {x - 2y} \right)^2} + {\left( {x - 2} \right)^2} = 10\\
Do:x;y \in Z\\
10 = 1 + 9\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
{\left( {x - 2y} \right)^2} = 1\\
{\left( {x - 2} \right)^2} = 9
\end{array} \right.\\
\left\{ \begin{array}{l}
{\left( {x - 2y} \right)^2} = 9\\
{\left( {x - 2} \right)^2} = 1
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 5;y = 2\\
x = 5;y = 3\\
x = - 1;y = - 1\\
x = - 1;y = 0\\
x = 3;y = 0\\
x = 3;y = 3\\
x = 1;y = - 1\\
x = 1;y = 2
\end{array} \right.\\
Vậy\,\left( {x;y} \right) \in \left\{ \begin{array}{l}
\left( {5;2} \right);\left( {5;3} \right);\left( { - 1; - 1} \right);\left( { - 1;0} \right)\\
\left( {3;0} \right);\left( {3;3} \right);\left( {1; - 1} \right);\left( {1;2} \right)
\end{array} \right\}
\end{array}$
