Đáp án:

$\begin{array}{l}
1)a)Ư\left( { - 15} \right) = \left\{ { - 15; - 5; - 3; - 1;1;3;5;15} \right\}\\
b)B\left( { - 6} \right) = \left\{ { - 6k/k \in Z} \right\}\\
c) - 5 \le x < 20\\
 \Rightarrow x \in \left\{ { - 5; - 3; - 1;1;3;5;6;10;15} \right\}\\
2)a)3n + 2\\
 = 3n - 3 + 5\\
 = 3\left( {n - 1} \right) + 5\\
Do:3\left( {n - 1} \right) \vdots \left( {n - 1} \right)\\
 \Rightarrow 5 \vdots \left( {n - 1} \right)\\
 \Rightarrow \left( {n - 1} \right) \in \left\{ { - 5; - 1;1;5} \right\}\\
 \Rightarrow n \in \left\{ { - 4;0;2;6} \right\}\\
b){n^2} - n + 3\\
 = n\left( {n - 1} \right) + 3\\
 \Rightarrow 3 \vdots \left( {n - 1} \right)\\
 \Rightarrow \left( {n - 1} \right) \in \left\{ { - 3; - 1;1;3} \right\}\\
 \Rightarrow n \in \left\{ { - 2;0;2;4} \right\}\\
d)5n - 4 \vdots \left( {2n - 1} \right)\\
 \Rightarrow 2\left( {5n - 4} \right) \vdots \left( {2n - 1} \right)\\
 \Rightarrow 10n - 8 \vdots \left( {2n - 1} \right)\\
 \Rightarrow 5.\left( {2n - 1} \right) - 3 \vdots \left( {2n - 1} \right)\\
 \Rightarrow 3 \vdots \left( {2n - 1} \right)\\
 \Rightarrow 2n - 1 \in \left\{ { - 3; - 1;1;3} \right\}\\
 \Rightarrow n \in \left\{ { - 1;0;1;2} \right\}\\
B3)a)\left( {5x - 1} \right).\left( {y - 4} \right) = 4\\
 \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
5x - 1 = 1\\
y - 4 = 4
\end{array} \right.\\
\left\{ \begin{array}{l}
5x - 1 = 4\\
y - 4 = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
5x - 1 =  - 1\\
y - 4 =  - 4
\end{array} \right.\\
\left\{ \begin{array}{l}
5x - 1 =  - 4\\
y - 4 =  - 1
\end{array} \right.\\
5x - 1 = y - 4 = 2\\
5x - 1 = y - 4 =  - 2
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = \dfrac{2}{5}\left( {ktm} \right)\\
x = 1;y = 5\\
x = 0;y = 0\\
x =  - \dfrac{3}{5}\left( {ktm} \right)\\
x = \dfrac{3}{5}\left( {ktm} \right)\\
x =  - \dfrac{1}{5}\left( {ktm} \right)
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( {1;5} \right);\left( {0;0} \right)} \right\}\\
b)xy + x + 2y = 5\\
 \Rightarrow x.\left( {y + 1} \right) + 2y + 2 = 7\\
 \Rightarrow x\left( {y + 1} \right) + 2\left( {y + 1} \right) = 7\\
 \Rightarrow \left( {y + 1} \right)\left( {x + 2} \right) = 7\\
 \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
y + 1 = 1\\
x + 2 = 7
\end{array} \right.\\
\left\{ \begin{array}{l}
y + 1 =  - 1\\
x + 2 =  - 7
\end{array} \right.\\
\left\{ \begin{array}{l}
y + 1 = 7\\
x + 2 = 1
\end{array} \right.\\
\left\{ \begin{array}{l}
y + 1 =  - 7\\
x + 2 =  - 1
\end{array} \right.
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
y = 0;x = 5\\
y =  - 2;x =  - 9\\
y = 6;x =  - 1\\
y =  - 8;x =  - 3
\end{array} \right.\\
Vậy\,\left( {x;y} \right) = \left\{ {\left( {5;0} \right);\left( { - 9; - 2} \right);\left( { - 1;6} \right);\left( { - 3; - 8} \right)} \right\}
\end{array}$

$\begin{array}{l} 1)a)Ư\left( { - 15} \right) = \left\{ { - 15; - 5; - 3; - 1;1;3;5;15} \right\}\\ b)B\left( { - 6} \right) = \left\{ { - 6k/k \in Z} \right\}\\ c) - 5 \le x < 20\\  \Rightarrow x \in \left\{ { - 5; - 3; - 1;1;3;5;6;10;15} \right\}\\ 2)a)3n + 2\\  = 3n - 3 + 5\\  = 3\left( {n - 1} \right) + 5\\ Do:3\left( {n - 1} \right) \vdots \left( {n - 1} \right)\\  \Rightarrow 5 \vdots \left( {n - 1} \right)\\  \Rightarrow \left( {n - 1} \right) \in \left\{ { - 5; - 1;1;5} \right\}\\  \Rightarrow n \in \left\{ { - 4;0;2;6} \right\}\\ b){n^2} - n + 3\\  = n\left( {n - 1} \right) + 3\\  \Rightarrow 3 \vdots \left( {n - 1} \right)\\  \Rightarrow \left( {n - 1} \right) \in \left\{ { - 3; - 1;1;3} \right\}\\  \Rightarrow n \in \left\{ { - 2;0;2;4} \right\}\\ d)5n - 4 \vdots \left( {2n - 1} \right)\\  \Rightarrow 2\left( {5n - 4} \right) \vdots \left( {2n - 1} \right)\\  \Rightarrow 10n - 8 \vdots \left( {2n - 1} \right)\\  \Rightarrow 5.\left( {2n - 1} \right) - 3 \vdots \left( {2n - 1} \right)\\  \Rightarrow 3 \vdots \left( {2n - 1} \right)\\  \Rightarrow 2n - 1 \in \left\{ { - 3; - 1;1;3} \right\}\\  \Rightarrow n \in \left\{ { - 1;0;1;2} \right\}\\ B3)a)\left( {5x - 1} \right).\left( {y - 4} \right) = 4\\  \Rightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} 5x - 1 = 1\\ y - 4 = 4 \end{array} \right.\\ \left\{ \begin{array}{l} 5x - 1 = 4\\ y - 4 = 1 \end{array} \right.\\ \left\{ \begin{array}{l} 5x - 1 =  - 1\\ y - 4 =  - 4 \end{array} \right.\\ \left\{ \begin{array}{l} 5x - 1 =  - 4\\ y - 4 =  - 1 \end{array} \right.\\ 5x - 1 = y - 4 = 2\\ 5x - 1 = y - 4 =  - 2 \end{array} \right. \Rightarrow \left[ \begin{array}{l} x = \dfrac{2}{5}\left( {ktm} \right)\\ x = 1;y = 5\\ x = 0;y = 0\\ x =  - \dfrac{3}{5}\left( {ktm} \right)\\ x = \dfrac{3}{5}\left( {ktm} \right)\\ x =  - \dfrac{1}{5}\left( {ktm} \right) \end{array} \right.\\ Vậy\,\left( {x;y} \right) = \left\{ {\left( {1;5} \right);\left( {0;0} \right)} \right\}\\ b)xy + x + 2y = 5\\  \Rightarrow x.\left( {y + 1} \right) + 2y + 2 = 7\\  \Rightarrow x\left( {y + 1} \right) + 2\left( {y + 1} \right) = 7\\  \Rightarrow \left( {y + 1} \right)\left( {x + 2} \right) = 7\\  \Rightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} y + 1 = 1\\ x + 2 = 7 \end{array} \right.\\ \left\{ \begin{array}{l} y + 1 =  - 1\\ x + 2 =  - 7 \end{array} \right.\\ \left\{ \begin{array}{l} y + 1 = 7\\ x + 2 = 1 \end{array} \right.\\ \left\{ \begin{array}{l} y + 1 =  - 7\\ x + 2 =  - 1 \end{array} \right. \end{array} \right. \Rightarrow \left[ \begin{array}{l} y = 0;x = 5\\ y =  - 2;x =  - 9\\ y = 6;x =  - 1\\ y =  - 8;x =  - 3 \end{array} \right.\\ Vậy\,\left( {x;y} \right) = \left\{ {\left( {5;0} \right);\left( { - 9; - 2} \right);\left( { - 1;6} \right);\left( { - 3; - 8} \right)} \right\} \end{array}$