Đáp án:
$\begin{array}{l}
a)\dfrac{{63}}{{3n + 1}} \in Z\\
\Rightarrow \left( {3n + 1} \right) \in U\left( {63} \right)\\
\Rightarrow \left( {3n + 1} \right) \in \left\{ \begin{array}{l}
- 63; - 21; - 9; - 7; - 3; - 1;\\
63;21;9;7;3;1
\end{array} \right\}\\
\Rightarrow 3n \in \left\{ \begin{array}{l}
- 64; - 22; - 10; - 8; - 4; - 2;\\
62;20;8;6;2;0
\end{array} \right\}\\
\Rightarrow n \in \left\{ {2;0} \right\}\left( {do:n \in Z} \right)\\
b)\dfrac{{{n^2} - 4n}}{{n - 3}}\\
= \dfrac{{{n^2} - 3n - n + 3 - 3}}{{n - 3}}\\
= \dfrac{{n\left( {n - 3} \right) - \left( {n - 3} \right) - 3}}{{n - 3}}\\
= \dfrac{{\left( {n - 3} \right)\left( {n - 1} \right) - 3}}{{n - 3}}\\
= n - 1 - \dfrac{3}{{n - 3}}\\
\Rightarrow 3 \vdots \left( {n - 3} \right)\\
\Rightarrow n - 3 \in \left\{ { - 3; - 1;1;3} \right\}\\
\Rightarrow n \in \left\{ {0;2;4;6} \right\}
\end{array}$
