Đáp án:
$\begin{array}{l}
a)\dfrac{{13}}{{n - 1}} \in Z\\
\Rightarrow \left( {n - 1} \right) \in U\left( {13} \right)\\
\Rightarrow \left( {n - 1} \right) \in \left\{ { - 13; - 1;1;13} \right\}\\
\Rightarrow n \in \left\{ { - 12;0;2;14} \right\}\\
b)\dfrac{n}{{5n + 1}} \in Z\\
\Rightarrow n = 0\\
c)\dfrac{{n - 2}}{5} \in Z\\
\Rightarrow \left( {n - 1} \right) \vdots 5\\
\Rightarrow n - 1 = 5.k\left( {k \in Z} \right)\\
\Rightarrow n = 5k + 1\left( {k \in Z} \right)\\
d)\dfrac{{ - 32}}{{2n - 3}} \in Z\\
\Rightarrow 2n - 3 \in U\left( {32} \right)\\
\Rightarrow \left( {2n - 3} \right) \in \left\{ \begin{array}{l}
- 32; - 16; - 8; - 4; - 2; - 1;\\
1;2;4;8;16;32
\end{array} \right\}\\
\Rightarrow 2n \in \left\{ { - 29; - 13; - 5; - 1;1;2;4;5;7;11;19;35} \right\}\\
\Rightarrow 2n \in \left\{ {2;4} \right\}\\
\Rightarrow n \in \left\{ {1;2} \right\}\\
e)\dfrac{{2n + 7}}{5} \in Z\\
\Rightarrow 2n + 7 = 5k\left( {k \in Z} \right)\\
\Rightarrow 2n = 5k - 7\\
\Rightarrow n = \dfrac{{5k - 7}}{2}\left( {k = 2a + 1\left( {a \in Z} \right)} \right)
\end{array}$
