Đáp án: $\left( {x;y} \right) = \left\{ {\left( {6;2} \right);\left( {4;2} \right);\left( {8;3} \right);\left( {2;1} \right)} \right\}$
Giải thích các bước giải:
$\begin{array}{l}
{x^2} + 10{y^2} - 6xy + 2x - 10y + 4 = 0\\
\Leftrightarrow {x^2} + 9{y^2} + 1 + 2x - 6xy - 6y\\
+ {y^2} - 4y + 4 = 1\\
\Leftrightarrow {\left( {x - 3y + 1} \right)^2} + {\left( {y - 2} \right)^2} = 1\\
Do:x;y \in Z\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
{\left( {x - 3y + 1} \right)^2} = 1\\
{\left( {y - 2} \right)^2} = 0
\end{array} \right.\\
\left\{ \begin{array}{l}
{\left( {x - 3y + 1} \right)^2} = 0\\
{\left( {y - 2} \right)^2} = 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
\left[ \begin{array}{l}
x = 3y\\
x = 3y - 2
\end{array} \right.\\
y = 2
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 3y - 1\\
\left[ \begin{array}{l}
y = 3\\
y = 1
\end{array} \right.
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 6;y = 2\\
x = 4;y = 2\\
x = 8;y = 3\\
x = 2;y = 1
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( {6;2} \right);\left( {4;2} \right);\left( {8;3} \right);\left( {2;1} \right)} \right\}
\end{array}$
