Đáp án:
$\begin{array}{l}
A = \dfrac{{x - 2}}{3} \in Z\\
\Rightarrow \left( {x - 2} \right) \vdots 3\\
\Rightarrow \left( {x - 2} \right) \in B\left( 3 \right)\\
\Rightarrow x - 2 = 3.k\left( {k \in Z} \right)\\
\Rightarrow x = 3k + 2\left( {k \in Z} \right)\\
\text{Vậy}\,x = 3k + 2\left( {k \in Z} \right)\\
B = \dfrac{5}{{x + 3}}\left( { \in Z} \right)\\
\Rightarrow 5 \vdots \left( {x + 3} \right)\\
\Rightarrow \left( {x + 3} \right) \in \left\{ { - 5; - 1;1;5} \right\}\\
\Rightarrow x \in \left\{ { - 8; - 4; - 2;2} \right\}\\
\text{Vậy}\,x \in \left\{ { - 8; - 4; - 2;2} \right\}\\
C = \dfrac{{x + 1}}{{x - 2}} = \dfrac{{x - 2 + 3}}{{x - 2}}\\
= \dfrac{{x - 2}}{{x - 2}} + \dfrac{3}{{x - 2}}\\
= 1 + \dfrac{3}{{x - 2}} \in Z\\
\Rightarrow \dfrac{3}{{x - 2}} \in Z\\
\Rightarrow \left( {x - 2} \right) \in \left\{ { - 3; - 1;1;3} \right\}\\
\Rightarrow x \in \left\{ { - 1;1;3;5} \right\}\\
\text{Vậy}\,x \in \left\{ { - 1;1;3;5} \right\}
\end{array}$
