Đáp án:

d. \(\left[ \begin{array}{l}
x = 7\\
x = 1\\
x = 5\\
x = 3
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
a.DK:x \ne  - \dfrac{1}{2}\\
A = \dfrac{2}{{2x + 1}}\\
A \in Z \Leftrightarrow \dfrac{2}{{2x + 1}} \in Z\\
 \Leftrightarrow 2x + 1 \in U\left( 2 \right)\\
 \Leftrightarrow \left[ \begin{array}{l}
2x + 1 = 2\\
2x + 1 =  - 2\\
2x + 1 = 1\\
2x + 1 =  - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = \dfrac{1}{2}\left( l \right)\\
x =  - \dfrac{3}{2}\left( l \right)\\
x = 0\left( {TM} \right)\\
x =  - 1\left( {TM} \right)
\end{array} \right.\\
b.DK:x \ne  - \dfrac{1}{2}\\
B =  - \dfrac{3}{{2x + 1}}\\
B \in Z \Leftrightarrow \dfrac{3}{{2x + 1}} \in Z\\
 \Leftrightarrow 2x + 1 \in U\left( 3 \right)\\
 \to \left[ \begin{array}{l}
2x + 1 = 3\\
2x + 1 =  - 3\\
2x + 1 = 1\\
2x + 1 =  - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 1\left( {TM} \right)\\
x =  - 2\left( {TM} \right)\\
x = 0\left( {TM} \right)\\
x =  - 1\left( {TM} \right)
\end{array} \right.\\
c.DK:x \ne 2\\
C = \dfrac{{x + 1}}{{x - 2}} = \dfrac{{x - 2 + 3}}{{x - 2}} = 1 + \dfrac{3}{{x - 2}}\\
C \in Z \Leftrightarrow \dfrac{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\\
x =  - 1\\
x = 3\\
x = 1
\end{array} \right.\\
d.DK:x \ne 4\\
D = \dfrac{{3\left( {x - 4} \right) + 3}}{{x - 4}} = 3 + \dfrac{3}{{x - 4}}\\
D \in Z \to \dfrac{3}{{x - 4}} \in Z\\
 \to x - 4 \in U\left( 3 \right)\\
 \to \left[ \begin{array}{l}
x - 4 = 3\\
x - 4 =  - 3\\
x - 4 = 1\\
x - 4 =  - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 7\\
x = 1\\
x = 5\\
x = 3
\end{array} \right.\\
e.DK:x \ne \dfrac{1}{2}\\
E = \dfrac{{6x + 5}}{{2x - 1}} = \dfrac{{3\left( {2x - 1} \right) + 8}}{{2x - 1}} = 3 + \dfrac{8}{{2x - 1}}\\
E \in Z \Leftrightarrow \dfrac{8}{{2x - 1}} \in Z\\
 \Leftrightarrow 2x - 1 \in U\left( 8 \right)\\
 \to \left[ \begin{array}{l}
2x - 1 = 8\\
2x - 1 =  - 8\\
2x - 1 = 4\\
2x - 1 =  - 4\\
2x - 1 = 2\\
2x - 1 =  - 2\\
2x - 1 = 1\\
2x - 1 =  - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = \dfrac{9}{2}\left( l \right)\\
x =  - \dfrac{7}{2}\left( l \right)\\
x = \dfrac{5}{2}\left( l \right)\\
x =  - \dfrac{3}{2}\left( l \right)\\
x = \dfrac{3}{2}\left( l \right)\\
x =  - \dfrac{1}{2}\left( l \right)\\
x = 1\left( {TM} \right)\\
x = 0\left( {TM} \right)
\end{array} \right.
\end{array}\)