Đáp án:

a) \(\left[ \begin{array}{l}
x = 3\\
x =  - 1\\
x = 2\\
x = 0
\end{array} \right.\) 

Giải thích các bước giải:

\(\begin{array}{l}
a)\dfrac{2}{{x - 1}} \in Z\\
 \Leftrightarrow x - 1 \in U\left( 2 \right)\\
 \to \left[ \begin{array}{l}
x - 1 = 2\\
x - 1 =  - 2\\
x - 1 = 1\\
x - 1 =  - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 3\\
x =  - 1\\
x = 2\\
x = 0
\end{array} \right.\\
b) - \dfrac{6}{{3x - 2}} \in Z\\
 \Leftrightarrow 3x - 2 \in U\left( 6 \right)\\
 \to \left[ \begin{array}{l}
3x - 2 = 6\\
3x - 2 =  - 6\\
3x - 2 = 3\\
3x - 2 =  - 3\\
3x - 2 = 2\\
3x - 2 =  - 2\\
3x - 2 = 1\\
3x - 2 =  - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = \dfrac{8}{3}\left( l \right)\\
x =  - \dfrac{4}{3}\left( l \right)\\
x = \dfrac{5}{3}\left( l \right)\\
x =  - \dfrac{1}{3}\left( l \right)\\
x = \dfrac{4}{3}\left( l \right)\\
x = 0\left( {TM} \right)\\
x = 1\left( {TM} \right)\\
x = \dfrac{1}{3}\left( l \right)
\end{array} \right.\\
c)\dfrac{{x - 2}}{{x - 1}} = \dfrac{{x - 1 - 1}}{{x - 1}} = 1 - \dfrac{1}{{x - 1}}\\
\dfrac{{x - 2}}{{x - 1}} \in Z\\
 \Leftrightarrow \dfrac{1}{{x - 1}} \in Z\\
 \Leftrightarrow x - 1 \in U\left( 1 \right)\\
 \to \left[ \begin{array}{l}
x - 1 = 1\\
x - 1 =  - 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 2\\
x = 0
\end{array} \right.\\
d)\dfrac{{2x + 3}}{{x - 5}} = \dfrac{{2\left( {x - 5} \right) + 13}}{{x - 5}} = 2 + \dfrac{{13}}{{x - 5}}\\
\dfrac{{2x + 3}}{{x - 5}} \in Z \Leftrightarrow \dfrac{{13}}{{x - 5}} \in Z\\
 \Leftrightarrow x - 5 \in U\left( {13} \right)\\
 \to \left[ \begin{array}{l}
x - 5 = 13\\
x - 5 =  - 13\\
x - 5 = 1\\
x - 5 =  - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 18\\
x =  - 8\\
x = 6\\
x = 4
\end{array} \right.
\end{array}\)