Đáp án:
\(\left[ \begin{array}{l}
x = 5\left( {TM} \right)\\
x = - 1\left( {TM} \right)\\
x = 3\left( {TM} \right)\\
x = 1\left( {TM} \right)
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
A = \frac{{x + 1}}{{x - 2}} = \frac{{x - 2 + 3}}{{x - 2}} = 1 + \frac{3}{{x - 2}}\left( {x \ne 2} \right)\\
Do:A \in Z\\
\to \frac{{x + 1}}{{x - 2}} \in Z\\
\to \frac{3}{{x - 2}} \in Z\\
\to x - 2 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
x - 2 = 3\\
x - 2 = - 3\\
x - 2 = 1\\
x - 2 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 5\left( {TM} \right)\\
x = - 1\left( {TM} \right)\\
x = 3\left( {TM} \right)\\
x = 1\left( {TM} \right)
\end{array} \right.
\end{array}\)
