Đáp án:
d. \(\left[ \begin{array}{l}
x = 2\\
x = - 4
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a.A = \dfrac{{x + 1}}{{x - 2}} = \dfrac{{x - 2 + 3}}{{x - 2}} = 1 + \dfrac{3}{{x - 2}}\\
A \in Z\\
\Leftrightarrow \dfrac{3}{{x - 2}} \in Z\\
\Leftrightarrow x - 2 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
x - 2 = 3\\
x - 2 = - 3\\
x - 2 = 1\\
x - 2 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 5\\
x = - 1\\
x = 3\\
x = 1
\end{array} \right.\\
b.A = \dfrac{{x - 1}}{{x - 2}} = \dfrac{{x - 2 + 1}}{{x - 2}} = 1 + \dfrac{1}{{x - 2}}\\
A \in Z\\
\Leftrightarrow \dfrac{1}{{x - 2}} \in Z\\
\Leftrightarrow x - 2 \in U\left( 1 \right)\\
\to \left[ \begin{array}{l}
x - 2 = 1\\
x - 2 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = 1
\end{array} \right.\\
c.A = \dfrac{{x + 1}}{{x + 2}} = \dfrac{{x + 2 - 1}}{{x + 2}} = 1 - \dfrac{1}{{x + 2}}\\
A \in Z\\
\Leftrightarrow \dfrac{1}{{x + 2}} \in Z\\
\to x + 2 \in U\left( 1 \right)\\
\to \left[ \begin{array}{l}
x + 2 = 1\\
x + 2 = - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = - 1\\
x = - 3
\end{array} \right.\\
d.A = \dfrac{{2x - 1}}{{x + 1}} = \dfrac{{2\left( {x + 1} \right) - 3}}{{x + 1}} = 2 - \dfrac{3}{{x + 1}}\\
A \in Z\\
\Leftrightarrow \dfrac{3}{{x + 1}} \in Z\\
\Leftrightarrow x + 1 \in U\left( 3 \right)\\
\to \left[ \begin{array}{l}
x + 1 = 3\\
x + 1 = - 3
\end{array} \right. \to \left[ \begin{array}{l}
x = 2\\
x = - 4
\end{array} \right.
\end{array}\)
