Đáp án: $x \in \left\{ { - 7; - 1;1;3;5;11} \right\}$
Giải thích các bước giải:
$\begin{array}{l}
A = \dfrac{{2{x^2} + x - 1}}{{x - 2}}\\
= \dfrac{{2{x^2} - 4x + 5x - 10 + 9}}{{x - 2}}\\
= \dfrac{{2x\left( {x - 2} \right) + 5\left( {x - 2} \right) + 9}}{{x - 2}}\\
= \dfrac{{\left( {x - 2} \right)\left( {2x + 5} \right) + 9}}{{x - 2}}\\
= 2x + 5 + \dfrac{9}{{x - 2}}\\
A \in Z\\
\Leftrightarrow \dfrac{9}{{x - 2}} \in Z\\
\Leftrightarrow \left( {x - 2} \right) \in \left\{ { - 9; - 3; - 1;1;3;9} \right\}\\
\Leftrightarrow x \in \left\{ { - 7; - 1;1;3;5;11} \right\}\\
Vay\,x \in \left\{ { - 7; - 1;1;3;5;11} \right\}
\end{array}$
