Đáp án:
\(\left[ \begin{array}{l}
x = 9\\
x = - 2\\
x = 4\\
x = 3
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ne \dfrac{7}{2}\\
B = \dfrac{{x + 2}}{{2x - 7}}\\
\to 2B = \dfrac{{2x + 4}}{{2x - 7}} = \dfrac{{2x - 7 + 11}}{{2x - 7}}\\
= 1 + \dfrac{{11}}{{2x - 7}}\\
B \in Z \to \dfrac{{11}}{{2x - 7}} \in Z\\
\to 2x - 7 \in U\left( {11} \right)\\
\to \left[ \begin{array}{l}
2x - 7 = 11\\
2x - 7 = - 11\\
2x - 7 = 1\\
2x - 7 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 9\\
x = - 2\\
x = 4\\
x = 3
\end{array} \right.
\end{array}\)
