Đáp án:
$\begin{array}{l}
a)xy + 3x - 2y - 9 = 0\\
\Leftrightarrow x.\left( {y + 3} \right) - 2y - 6 - 3 = 0\\
\Leftrightarrow x.\left( {y + 3} \right) - 2.\left( {y + 3} \right) = 3\\
\Leftrightarrow \left( {y + 3} \right)\left( {x - 2} \right) = 3 = 1.3 = \left( { - 1} \right).\left( { - 3} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
y + 3 = 1\\
x - 2 = 3
\end{array} \right.\\
\left\{ \begin{array}{l}
y + 3 = 3\\
x - 2 = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
y + 3 = - 1\\
x - 2 = - 3
\end{array} \right.\\
\left\{ \begin{array}{l}
y + 3 = - 3\\
x - 2 = - 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
y = - 2;x = 5\\
y = 0;x = 3\\
y = - 4;x = - 1\\
y = - 6;x = 1
\end{array} \right.\\
Vậy\,\left( {y;x} \right) = \left\{ {\left( { - 2;5} \right);\left( {0;3} \right);\left( { - 4; - 1} \right);\left( { - 6;1} \right)} \right\}\\
b)xy - x + 3y - 4 = 0\\
\Leftrightarrow x\left( {y - 1} \right) + 3y - 3 - 1 = 0\\
\Leftrightarrow x\left( {y - 1} \right) + 3\left( {y - 1} \right) = 1\\
\Leftrightarrow \left( {y - 1} \right)\left( {x + 3} \right) = 1 = 1.1 = \left( { - 1} \right).\left( { - 1} \right)\\
\Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
y - 1 = 1\\
x + 3 = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
y - 1 = - 1\\
x + 3 = - 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
y = 2;x = - 2\\
y = 0;x = - 4
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( { - 2;2} \right);\left( { - 4;0} \right)} \right\}
\end{array}$
