Đáp án: $\left( {x;y} \right) = \left\{ {\left( {1; - 6} \right);\left( {3;6} \right);\left( {7;6} \right);\left( { - 3; - 6} \right)} \right\}$
Giải thích các bước giải:
$\begin{array}{l}
{x^2} - 4x + 2y - xy + 9 = 0\\
\Leftrightarrow {x^2} - 4x + 4 + y\left( {2 - x} \right) + 5 = 0\\
\Leftrightarrow {\left( {x - 2} \right)^2} - y\left( {x - 2} \right) = - 5\\
\Leftrightarrow \left( {x - 2} \right)\left( {x - 2 - y} \right) = - 5 = - 1.5 = \left( { - 5} \right).1\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x - 2 = - 1\\
x - 2 - y = 5
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 2 = 1\\
x - 2 - y = - 5
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 2 = 5\\
x - 2 - y = - 1
\end{array} \right.\\
\left\{ \begin{array}{l}
x - 2 = - 5\\
x - 2 - y = 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = 1\\
y = x - 7 = - 6
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 3\\
y = x + 3 = 6
\end{array} \right.\\
\left\{ \begin{array}{l}
x = 7\\
y = x - 1 = 6
\end{array} \right.\\
\left\{ \begin{array}{l}
x = - 3\\
y = x - 3 = - 6
\end{array} \right.
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( {1; - 6} \right);\left( {3;6} \right);\left( {7;6} \right);\left( { - 3; - 6} \right)} \right\}
\end{array}$
