Đáp án:
$\begin{array}{l}
a)\left( {3x + 2} \right) \vdots \left( {2x - 1} \right)\\
\Rightarrow 2\left( {3x + 2} \right) \vdots \left( {2x - 1} \right)\\
\Rightarrow 6x + 4 \vdots \left( {2x - 1} \right)\\
\Rightarrow 6x - 3 + 7 \vdots \left( {2x - 1} \right)\\
\Rightarrow 3\left( {2x - 1} \right) + 7 \vdots \left( {2x - 1} \right)\\
Do:3\left( {2x - 1} \right) \vdots \left( {2x - 1} \right)\\
\Rightarrow 7 \vdots \left( {2x - 1} \right)\\
\Rightarrow \left( {2x - 1} \right) \in Ư\left( 7 \right) = {\rm{\{ }} - 7; - 1;1;7\} \\
\Rightarrow x \in \left\{ { - 3;0;1;4} \right\}
\end{array}$
b) Do ${\left( {x - 5} \right)^2} \ge 0\forall x$ nên:
$\begin{array}{l}
{\left( {x - 5} \right)^2}.\left( {2x - 6} \right) < 0\\
\Rightarrow 2x - 6 < 0\\
\Rightarrow 2x < 6\\
\Rightarrow x < 3
\end{array}$
Vậy với mọi số nguyên x<3 thì thỏa mãn đề bài
