Đáp án:
b) \(\left[ \begin{array}{l}
x = 2\\
x = - 8\\
x = - 2\\
x = - 4
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
A = \dfrac{{x + 2}}{{x - 5}} = \dfrac{{x - 5 + 7}}{{x - 5}} = 1 + \dfrac{7}{{x - 5}}\\
A \in Z \to \dfrac{7}{{x - 5}} \in Z\\
\to x - 5 \in U\left( 7 \right)\\
\to \left[ \begin{array}{l}
x - 5 = 7\\
x - 5 = - 7\\
x - 5 = 1\\
x - 5 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 12\\
x = - 2\\
x = 6\\
x = 4
\end{array} \right.\\
D = \dfrac{3}{{x - 1}}\\
D \in Z \to \dfrac{3}{{x - 1}} \in Z\\
\to x - 1 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
x - 1 = 3\\
x - 1 = - 3\\
x - 1 = 1\\
x - 1 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 4\\
x = - 2\\
x = 2\\
x = 0
\end{array} \right.\\
C = \dfrac{{2x + 1}}{{x - 3}} = \dfrac{{2\left( {x - 3} \right) + 7}}{{x - 3}}\\
= 2 + \dfrac{7}{{x - 3}}\\
C \in Z \to \dfrac{7}{{x - 3}} \in Z\\
\to x - 3 \in U\left( 7 \right)\\
\to \left[ \begin{array}{l}
x - 3 = 7\\
x - 3 = - 7\\
x - 3 = 1\\
x - 3 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 10\\
x = - 4\\
x = 4\\
x = 2
\end{array} \right.\\
B = \dfrac{{x - 2}}{{x + 3}} = \dfrac{{x + 3 - 5}}{{x + 3}} = 1 - \dfrac{5}{{x + 3}}\\
B \in Z \to \dfrac{5}{{x + 3}} \in Z\\
\to x + 3 \in U\left( 5 \right)\\
\to \left[ \begin{array}{l}
x + 3 = 5\\
x + 3 = - 5\\
x + 3 = 1\\
x + 3 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 2\\
x = - 8\\
x = - 2\\
x = - 4
\end{array} \right.
\end{array}\)
