Đáp án:
$\begin{array}{l}
f\left( x \right) = 2{x^2} + x + 5\\
= 2{x^2} - x + 2x - 1 + 6\\
= x.\left( {2x - 1} \right) + \left( {2x - 1} \right) + 6\\
= \left( {2x - 1} \right).\left( {x + 1} \right) + 6\\
Do:\left( {2x - 1} \right)\left( {x + 1} \right) \vdots \left( {2x - 1} \right)\\
\Rightarrow 6 \vdots \left( {2x - 1} \right)\\
\Rightarrow \left( {2x - 1} \right) \in \left\{ { - 6; - 3; - 2; - 1;1;2;3;6} \right\}\\
\Rightarrow 2x \in \left\{ { - 5; - 2; - 1;0;2;3;4;7} \right\}\\
\Rightarrow x \in \left\{ { - \frac{5}{2}; - 1; - \frac{1}{2};0;1;\frac{3}{2};2;\frac{7}{2}} \right\}\\
Do:x \in Z\\
\Rightarrow x \in \left\{ { - 1;0;1;2} \right\}\\
Vậy\,x \in \left\{ { - 1;0;1;2} \right\}
\end{array}$
