Giải thích các bước giải:
\(\begin{array}{l}
a)\,\left( {2n + 5} \right)\, \vdots \,\left( {n - 2} \right)\\
\left( {2\left( {n - 2} \right) + 9} \right)\, \vdots \,\left( {n - 2} \right)\\
9\, \vdots \,\left( {n - 2} \right)\,hay\,\left( {n - 2} \right) \in U\left( 9 \right) = \left\{ { - 1;1; - 3;3; - 9;9} \right\}\\
Suy\,ra\,n \in \left\{ {1;3; - 1;5; - 7;11} \right\}\\
1)\,a)\,\,A \in Z\\
nen\,\left( {n + 6} \right)\, \vdots \,\left( {n + 2} \right)\\
\left( {n + 2 + 4} \right)\, \vdots \,\left( {n + 2} \right)\\
4\, \vdots \,\left( {n + 2} \right)\,hay\,\left( {n + 2} \right) \in U\left( 4 \right) = \left\{ { - 1;1; - 2;2; - 4;4} \right\}\\
Suy\,ra\,n \in \left\{ { - 3; - 1; - 4;0;2; - 6} \right\}\\
b)\,B \in Z\\
nen\,\left( {2n + 5} \right)\, \vdots \,\left( {n + 3} \right)\\
\left( {2\left( {n + 3} \right) - 1} \right)\, \vdots \,\left( {n + 3} \right)\\
1\, \vdots \,\left( {n + 3} \right)\,hay\,\left( {n + 3} \right) \in U\left( 1 \right) = \left\{ { - 1;1} \right\}\\
Suy\,ra\,n \in \left\{ { - 4; - 2} \right\}\\
c)\,C \in Z \Rightarrow 2C \in Z \Rightarrow \dfrac{{6n + 22}}{{2n + 3}} \in Z\\
\Rightarrow \dfrac{{3\left( {2n + 3} \right) + 13}}{{2n + 3}} \in Z\\
3 + \dfrac{{13}}{{2n + 3}} \in Z \Rightarrow \left( {2n + 3} \right) \in U\left( {13} \right) = \left\{ { - 1;1;13; - 13} \right\}\\
\Rightarrow n \in \left\{ { - 2; - 1;5; - 8} \right\}
\end{array}\)
