Giải thích các bước giải:
\(\begin{array}{l}
a.\left( {x + 5} \right)\left( {3x - 12} \right) > 0\\
\to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x + 5 > 0\\
3x - 12 > 0
\end{array} \right.\\
\left\{ \begin{array}{l}
x + 5 < 0\\
3x - 12 < 0
\end{array} \right.
\end{array} \right. \to \left[ \begin{array}{l}
\left\{ \begin{array}{l}
x > - 5\\
x > 4
\end{array} \right.\\
\left\{ \begin{array}{l}
x < - 5\\
x < 4
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
x > 4\\
x < - 5
\end{array} \right.\\
b.\left\{ \begin{array}{l}
xy + 3x - 7y = 21{\rm{ }}\\
xy + 3x - 2y = 11
\end{array} \right. \to \left\{ \begin{array}{l}
5y = - 10\\
xy + 3x - 2y = 11
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = - 2\\
x = 7
\end{array} \right.\\
c.\left( {3n - 5} \right) \vdots n\\
\Leftrightarrow 5 \vdots n\\
\Leftrightarrow n \in U\left( 5 \right)\\
\to \left[ \begin{array}{l}
n = \pm 5\\
n = \pm 1
\end{array} \right.\\
d.\left( {n - 3 + 13} \right) \vdots \left( {n - 3} \right)\\
\Leftrightarrow 13 \vdots n - 3\\
\to n - 3 \in U\left( {13} \right)\\
\to \left[ \begin{array}{l}
n - 3 = 13\\
n - 3 = - 13\\
n - 3 = 1\\
n - 3 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
n = 16\\
n = - 10\\
n = 4\\
n = 2
\end{array} \right.
\end{array}\)