Đáp án:

 9) \(\left[ \begin{array}{l}
x = 2\\
x = 0
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
1)17 \vdots 2x + 3\\
 \Leftrightarrow 2x + 3 \in U\left( {17} \right)\\
 \to \left[ \begin{array}{l}
2x + 3 = 17\\
2x + 3 = 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 7\\
x =  - 1
\end{array} \right.\\
2)x + 12 \vdots x + 5\\
 \to x + 5 + 7 \vdots x + 5\\
 \to 7 \vdots x + 5\\
 \to x + 5 \in U\left( 7 \right)\\
 \to \left[ \begin{array}{l}
x + 5 = 7\\
x + 5 = 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 2\\
x =  - 4
\end{array} \right.\\
3)x - 1 \vdots x - 13\\
 \to x - 13 + 12 \vdots x - 13\\
 \to 12 \vdots x - 13\\
 \to x - 13 \in U\left( {12} \right)\\
 \to \left[ \begin{array}{l}
x - 13 = 12\\
x - 13 = 6\\
x - 13 = 4\\
x - 13 = 3\\
x - 13 = 2\\
x - 13 = 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 25\\
x = 19\\
x = 17\\
x = 16\\
x = 15\\
x = 14
\end{array} \right.\\
4)2x + 1 \in U\left( {2x + 23} \right)\\
 \to 2x + 23 \vdots 2x + 1\\
 \to 2x + 1 + 22 \vdots 2x + 1\\
 \to 22 \vdots 2x + 1\\
 \to 2x + 1 \in U\left( {22} \right)\\
 \to \left[ \begin{array}{l}
2x + 1 = 22\\
2x + 1 = 11\\
2x + 1 = 2\\
2x + 1 = 1
\end{array} \right. \to \left[ \begin{array}{l}
x = \dfrac{{21}}{2}\\
x = 5\\
x = \dfrac{1}{2}\\
x = 0
\end{array} \right.\\
5)x - 2 \in U\left( {2x} \right)\\
 \to 2x \vdots x - 2\\
 \to 2\left( {x - 2} \right) + 4 \vdots x - 2\\
 \to 4 \vdots x - 2\\
 \to x + 2 \in U\left( 4 \right)\\
 \to \left[ \begin{array}{l}
x + 2 = 4\\
x + 2 = 2\\
x + 2 = 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 2\\
x = 0\\
x =  - 1
\end{array} \right.\\
6)x + 10 \vdots 2x - 3\\
 \to 2x + 20 \vdots 2x - 3\\
 \to 2x - 3 + 23 \vdots 2x - 3\\
 \to 23 \vdots 2x - 3\\
 \to 2x - 3 \in U\left( {23} \right)\\
 \to \left[ \begin{array}{l}
2x - 3 = 23\\
2x - 3 = 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 13\\
x = 2
\end{array} \right.\\
7)2x + 1 \vdots x - 3\\
 \to 2\left( {x - 3} \right) + 7 \vdots x - 3\\
 \to 7 \vdots x - 3\\
 \to x - 3 \in U\left( 7 \right)\\
 \to \left[ \begin{array}{l}
x - 3 = 7\\
x - 3 = 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 10\\
x = 4
\end{array} \right.\\
8)x - 1 \in U\left( {3x + 25} \right)\\
 \to 3x + 25 \vdots x - 1\\
 \to 3\left( {x - 1} \right) + 28 \vdots x - 1\\
 \to 28 \vdots x - 1\\
 \to x - 1 \in U\left( {28} \right)\\
 \to \left[ \begin{array}{l}
x - 1 = 28\\
x - 1 = 14\\
x - 1 = 7\\
x - 1 = 4\\
x - 1 = 2\\
x - 1 = 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 29\\
x = 15\\
x = 8\\
x = 5\\
x = 3\\
x = 2
\end{array} \right.\\
9)2x + 3 \vdots 3x + 1\\
 \to 6x + 9 \vdots 3x + 1\\
 \to 2\left( {3x + 1} \right) + 7 \vdots 3x + 1\\
 \to 7 \vdots 3x + 1\\
 \to 3x + 1 \in U\left( 7 \right)\\
 \to \left[ \begin{array}{l}
3x + 1 = 7\\
3x + 1 = 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 2\\
x = 0
\end{array} \right.\\
10)5x + 4 \in U\left( {4x - 2} \right)\\
 \to 4x - 2 \vdots 5x + 4\\
 \to 20x - 10 \vdots 5x + 4\\
 \to 4\left( {5x + 4} \right) - 26 \vdots 5x + 4\\
 \to 26 \vdots 5x + 4\\
 \to \left[ \begin{array}{l}
5x + 4 = 26\\
5x + 4 = 13\\
5x + 4 = 2\\
5x + 4 = 1
\end{array} \right. \to \left[ \begin{array}{l}
x = \dfrac{{22}}{5}\\
x = \dfrac{9}{5}\\
x =  - \dfrac{2}{5}\\
x =  - \dfrac{3}{5}
\end{array} \right.
\end{array}\)