Đáp án:
$n \in \left\{ {0;1;2;3} \right\}$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
2{n^3} - 7{n^2} + 2n + 12\\
= {n^2}\left( {2n - 3} \right) - 2n\left( {2n - 3} \right) - 2\left( {2n - 3} \right) + 6\\
= \left( {2n - 3} \right)\left( {{n^2} - 2n - 2} \right) + 6
\end{array}$
Như vậy
$\begin{array}{l}
\left( {2{n^3} - 7{n^2} + 2n + 12} \right) \vdots \left( {2n - 3} \right)\\
\Leftrightarrow \left( {\left( {2n - 3} \right)\left( {{n^2} - 2n - 2} \right) + 6} \right) \vdots \left( {2n - 3} \right)\\
\Leftrightarrow 6 \vdots \left( {2n - 3} \right)\\
\Leftrightarrow \left( {2n - 3} \right) \in U\left( 6 \right) = \left\{ { - 6; - 3; - 2; - 1;1;2;3;6} \right\}
\end{array}$
Mặt khác $\left( {2n - 3} \right)\not \vdots 2,\forall n \in Z$
$\begin{array}{l}
\Rightarrow \left( {2n - 3} \right) \in \left\{ { - 3; - 1;1;3} \right\}\\
\Leftrightarrow n \in \left\{ {0;1;2;3} \right\}
\end{array}$
Vậy $n \in \left\{ {0;1;2;3} \right\}$ thỏa mãn đề.
