Giải thích các bước giải:
\(\begin{array}{l}
A = \frac{{4x + 1}}{{3x - 2}}\\
\to 3A = \frac{{12x + 3}}{{3x - 2}} = \frac{{4\left( {3x - 2} \right) + 11}}{{3x - 2}}
\end{array}\)
\( = 4 + \frac{{11}}{{3x - 2}}\)
Để A nguyên ⇔ 3A nguyên
⇔ \(\frac{{11}}{{3x - 2}}\) nguyên
\(\begin{array}{l}
\Leftrightarrow \left( {3x - 2} \right) \in U\left( {11} \right)\\
\to \left[ \begin{array}{l}
3x - 2 = 11\\
3x - 2 = - 11\\
3x - 2 = 1\\
3x - 2 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = \frac{{13}}{3}\left( l \right)\\
x = - 3\left( {TM} \right)\\
x = 1\left( {TM} \right)\\
x = \frac{1}{3}\left( l \right)
\end{array} \right.
\end{array}\)
