Đáp án:

\(\begin{array}{l}
b)\left[ \begin{array}{l}
x = 25\\
x =  - 35\\
x = 10\\
x =  - 20\\
x = 1\\
x =  - 11\\
x = 0\\
x =  - 10\\
x =  - 3\\
x =  - 7\\
x =  - 4\\
x =  - 6
\end{array} \right.\\
c)\left[ \begin{array}{l}
x = 6\\
x =  - 1\\
x = 3\\
x = 2
\end{array} \right.\\
d)\left[ \begin{array}{l}
x =  - 1\\
x =  - 3
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
C2:\\
a)A = \dfrac{{3\left( {x + 4} \right) - 14}}{{x + 4}}\\
 = 3 - \dfrac{{14}}{{x + 4}}\\
A \in Z \to \dfrac{{14}}{{x + 4}} \in Z\\
 \to x + 4 \in U\left( 4 \right)\\
 \to \left[ \begin{array}{l}
x + 4 = 14\\
x + 4 =  - 14\\
x + 4 = 7\\
x + 4 =  - 7\\
x + 4 = 2\\
x + 4 =  - 2\\
x + 4 = 1\\
x + 4 =  - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 10\\
x =  - 18\\
x = 3\\
x =  - 11\\
x =  - 2\\
x =  - 6\\
x =  - 3\\
x =  - 5
\end{array} \right.\\
b)B = \dfrac{{4\left( {x + 5} \right) - 30}}{{x + 5}}\\
 = 4 - \dfrac{{30}}{{x + 5}}\\
B \in Z \to \dfrac{{30}}{{x + 5}} \in Z\\
 \to x + 5 \in U\left( {30} \right)\\
 \to \left[ \begin{array}{l}
x + 5 = 30\\
x + 5 =  - 30\\
x + 5 = 15\\
x + 5 =  - 15\\
x + 5 = 6\\
x + 5 =  - 6\\
x + 5 = 5\\
x + 5 =  - 5\\
x + 5 = 2\\
x + 5 =  - 2\\
x + 5 = 1\\
x + 5 =  - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 25\\
x =  - 35\\
x = 10\\
x =  - 20\\
x = 1\\
x =  - 11\\
x = 0\\
x =  - 10\\
x =  - 3\\
x =  - 7\\
x =  - 4\\
x =  - 6
\end{array} \right.\\
c)C = \dfrac{{3x - 4}}{{2x - 5}}\\
 \to 2C = \dfrac{{6x - 8}}{{2x - 5}} = \dfrac{{3\left( {2x - 5} \right) + 7}}{{2x - 5}}\\
 = 3 + \dfrac{7}{{2x - 5}}\\
C \in Z \to \dfrac{7}{{2x - 5}} \in Z\\
 \to 2x - 5 \in U\left( 7 \right)\\
 \to \left[ \begin{array}{l}
2x - 5 = 7\\
2x - 5 =  - 7\\
2x - 5 = 1\\
2x - 5 =  - 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 6\\
x =  - 1\\
x = 3\\
x = 2
\end{array} \right.\\
d)D = \dfrac{{{x^2} + 4x + 4 - x - 1}}{{x + 2}}\\
 = \dfrac{{{{\left( {x + 2} \right)}^2} - \left( {x + 2} \right) + 1}}{{x + 2}}\\
 = \left( {x + 2} \right) - 1 + \dfrac{1}{{x + 2}}\\
D \in Z \to \dfrac{1}{{x + 2}} \in Z \to x + 2 \in U\left( 1 \right)\\
 \to \left[ \begin{array}{l}
x + 2 = 1\\
x + 2 =  - 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x =  - 1\\
x =  - 3
\end{array} \right.
\end{array}\)