Đáp án:
$\begin{array}{l}
\left( {{a^2} + a + 2} \right) \vdots \left( {a + 1} \right)\\
Do:{a^2} + a + 2 = a\left( {a + 1} \right) + 2\\
a\left( {a + 1} \right) \vdots \left( {a + 1} \right)\\
\Rightarrow 2 \vdots \left( {a + 1} \right)\\
\Rightarrow \left( {a + 1} \right) \in Ư\left( 2 \right) = {\rm{\{ }} - 2; - 1;1;2\} \\
\Rightarrow a \in {\rm{\{ }} - 3; - 2;0;1\}
\end{array}$
