Đáp án:
\[\left( {x;y} \right) = \left\{ {\left( { - 14;0} \right);\left( { - 2;6} \right)} \right\}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{x^2}y - x + xy = 14\\
\Leftrightarrow \left( {{x^2}y + xy} \right) - \left( {x + 1} \right) = 13\\
\Leftrightarrow xy\left( {x + 1} \right) - \left( {x + 1} \right) = 13\\
\Leftrightarrow \left( {xy - 1} \right)\left( {x + 1} \right) = 13\\
13 = 1.13 = \left( { - 1} \right).\left( { - 13} \right)\\
\Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
xy - 1 = 1\\
x + 1 = 13
\end{array} \right.\\
\left\{ \begin{array}{l}
xy - 1 = 13\\
x + 1 = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
xy - 1 = - 1\\
x + 1 = - 13
\end{array} \right.\\
\left\{ \begin{array}{l}
xy - 1 = - 13\\
x + 1 = - 1
\end{array} \right.
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x = 12\\
y = \frac{1}{6}
\end{array} \right.\left( L \right)\\
\left\{ \begin{array}{l}
x = 0\\
0y = 14
\end{array} \right.\left( L \right)\\
\left\{ \begin{array}{l}
x = - 14\\
y = 0
\end{array} \right.\left( {t/m} \right)\\
\left\{ \begin{array}{l}
x = - 2\\
y = 6
\end{array} \right.
\end{array} \right.\\
\Rightarrow \left( {x;y} \right) = \left\{ {\left( { - 14;0} \right);\left( { - 2;6} \right)} \right\}
\end{array}\)
