$\begin{array}{l}1)\, \dfrac{-13}{\sqrt{x}+1} \in \Bbb Z \, (x \geq 0)\\\Leftrightarrow(\sqrt{x}+1)\in Ư(-13)=\left\{\pm 1;\pm 13\right\}\\mà\, \sqrt{x}+1 \geq 1\\nên\, (\sqrt{x} + 1)= \left\{1;13\right\}\\\text{Ta có bảng giá trị:}\\\begin{array}{|l|r} \sqrt{x}+1 & 1& 13\\ \hline \sqrt{x} & 0& 12 \\ \hline x&0&144 \end{array}\\Vậy\, x = \left\{0;144\right\}\\2)\,\dfrac{\sqrt{x} - 5}{\sqrt{x} + 2}=\dfrac{\sqrt{x} + 2 - 7}{\sqrt{x} + 2}=1-\dfrac{7}{\sqrt{x}+2}\in \Bbb Z \, (x \geq 0)\\\Leftrightarrow(\sqrt{x}+2)\in Ư(7)=\left\{\pm 1; \pm 7\right\}\\Do \, \sqrt{x} + 2 \geq 2 \\nên \, \sqrt{x}+2=7\\\Leftrightarrow \sqrt{x} = 5\\\Leftrightarrow x = 25\\Vậy \, x = 25\\3)\,\dfrac{4\sqrt{x}+1}{2\sqrt{x}-3}=\dfrac{2(2\sqrt{x}-3)+7}{2\sqrt{x}-3}=2+\dfrac{7}{2\sqrt{x}-3}\in \Bbb Z \,(x \geq 0; \, x \in \Bbb Z)\\\Leftrightarrow (2\sqrt{x}-3)\in Ư(7)=\left\{\pm 1; \pm 7\right\}\\\text{Ta có bảng giá trị:}\\\begin{array}{|l|r} 2\sqrt{x}-3 & -7& -1&1&7\\ \hline \sqrt{x} & -2& 1&2&5 \\ \hline x&(loại)&1&4&25 \end{array}\\Vậy \, x=\left\{1;4;25\right\}\\4)\, \dfrac{2\sqrt{x}+6}{\sqrt{x}+5}=\dfrac{2(\sqrt{x} + 5) - 4}{\sqrt{x}+5}=2 -\dfrac{4}{\sqrt{x}+5}\in \Bbb Z \, (x\geq 0)\\\Leftrightarrow (\sqrt{x} + 5)\in Ư(4)=\left\{\pm 1;\pm 2;\, \pm4\right\}\\Do \, \sqrt{x} + 5 \geq 5 \\\text{nên không có x thỏa mãn}\\5)\, \dfrac{7-\sqrt{x}}{\sqrt{x}+8}=\dfrac{-(\sqrt{x}+8)+15}{\sqrt{x}+8}=-2+\dfrac{15}{\sqrt{x}+8}\in \Bbb Z \, (x \geq 0)\\\Leftrightarrow (\sqrt{x}+8)\in Ư(15)=\left\{\pm1; \pm3; \pm 5; \pm 15\right\}\\Do, \sqrt{x}+8\geq 8\\nên \, \sqrt{x}+8 = 15\\\Leftrightarrow \sqrt{x} = 7\\\Leftrightarrow x = 49\\\text{Vậy x = 49}\\6)\, \dfrac{3\sqrt{x}-2}{3\sqrt{x}-4}=\dfrac{3\sqrt{x} - 4 + 2}{3\sqrt{x} - 4}=1+\dfrac{2}{3\sqrt{x}-4}\in \Bbb Z \, (x\geq 0; \, x \in \Bbb Z)\\\Leftrightarrow (3\sqrt{x}-4)\in Ư(2)=\left\{\pm1; \pm2\right\}\\\text{Ta có bảng giá trị:}\\\begin{array}{|l|r} 3\sqrt{x}-4 & -2& -1&1&2\\ \hline \sqrt{x} & \dfrac{2}{3}& 1&\dfrac{5}{3}&2 \\ \hline x&(loại)&1&(loại)&4 \end{array}\\\text{Vậy x = {1;4}}\\7)\,\dfrac{3\sqrt{x}}{\sqrt{x}-1}=\dfrac{3(\sqrt{x}-1)+3}{\sqrt{x}-1}=3+\dfrac{3}{\sqrt{x}-1}\in \Bbb Z \, (x \geq 0)\\\Leftrightarrow (\sqrt{x} - 1)\in Ư(3)=\left\{\pm 1; \pm3\right\}\\Do\, \sqrt{x}-1\geq -1\\nên\, (\sqrt{x}-1)=\left\{\pm1;3\right\}\\\text{Ta có bảng giá trị:}\\\begin{array}{|l|r} \sqrt{x}-1 & -1& 1&3\\ \hline \sqrt{x} & 0& 2&4\\ \hline x&0&4&16 \end{array}\\\text{Vậy x = {0;4;16}}\\8)\, \dfrac{3-x}{\sqrt{x}+3}=\dfrac{(3-\sqrt{x})(\sqrt{x}+3)-6}{\sqrt{x}+3}=3-\sqrt{x}-\dfrac{6}{\sqrt{x}+3}\in \Bbb Z \, (x\geq 0)\\\Leftrightarrow (\sqrt{x}+3)\in Ư(6)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\\Do \, \sqrt{x} + 3 \geq 3\\nên\, (\sqrt{x}+3)=\left\{3;6\right\}\\\text{Ta có bảng giá trị:}\\\begin{array}{|l|r} \sqrt{x}+3 & 3& 6\\ \hline \sqrt{x} & 0& 3\\ \hline x&0&9 \end{array}\\\text{Vậy x = {0;9}}\end{array}$

Đáp án:

7) \(\left[ \begin{array}{l}
x = 16\\
x = 4\\
x = 0
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
1)DK:x \ge 0\\
B =  - \dfrac{{13}}{{\sqrt x  + 1}}\\
B \in Z\\
 \Leftrightarrow \dfrac{{13}}{{\sqrt x  + 1}} \in Z\\
 \to \sqrt x  + 1 \in U\left( {13} \right)\\
 \to \left[ \begin{array}{l}
\sqrt x  + 1 = 13\\
\sqrt x  + 1 = 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\sqrt x  = 12\\
\sqrt x  = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 144\\
x = 0
\end{array} \right.\\
2)DK:x \ge 0\\
C = \dfrac{{\sqrt x  - 5}}{{\sqrt x  + 2}} = \dfrac{{\sqrt x  + 2 - 7}}{{\sqrt x  + 2}}\\
 = 1 - \dfrac{7}{{\sqrt x  + 2}}\\
C \in Z\\
 \to \dfrac{7}{{\sqrt x  + 2}} \in Z\\
 \to \sqrt x  + 2 \in U\left( 7 \right)\\
 \to \left[ \begin{array}{l}
\sqrt x  + 2 = 7\\
\sqrt x  + 2 = 1\left( l \right)
\end{array} \right.\\
 \to \sqrt x  = 5\\
 \to x = 25\\
3)D = \dfrac{{4\sqrt x  + 1}}{{2\sqrt x  - 3}} = \dfrac{{2\left( {2\sqrt x  - 3} \right) + 7}}{{2\sqrt x  - 3}}\\
 = 2 + \dfrac{7}{{2\sqrt x  - 3}}\\
D \in Z\\
 \to \dfrac{7}{{2\sqrt x  - 3}} \in Z\\
 \to 2\sqrt x  - 3 \in U\left( 7 \right)\\
 \to \left[ \begin{array}{l}
2\sqrt x  - 3 = 7\\
2\sqrt x  - 3 = 1\\
2\sqrt x  - 3 =  - 1\\
2\sqrt x  - 3 =  - 7\left( l \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
2\sqrt x  = 10\\
2\sqrt x  = 4\\
2\sqrt x  = 2
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 25\\
x = 4\\
x = 1
\end{array} \right.\\
4)E = \dfrac{{2\sqrt x  + 6}}{{\sqrt x  + 5}} = \dfrac{{2\left( {\sqrt x  + 5} \right) - 4}}{{\sqrt x  + 5}}\\
 = 2 - \dfrac{4}{{\sqrt x  + 5}}\\
E \in Z\\
 \to \dfrac{4}{{\sqrt x  + 5}} \in Z\\
 \to \sqrt x  + 5 \in U\left( 4 \right)\left( {voly} \right)\\
Do:\sqrt x  + 5 \ge 5\forall x \ge 0\\
 \to x \in \emptyset \\
5)F = \dfrac{{7 - \sqrt x }}{{\sqrt x  + 8}} = \dfrac{{ - \left( {\sqrt x  - 7} \right)}}{{\sqrt x  + 8}}\\
 =  - \dfrac{{\sqrt x  + 8 - 15}}{{\sqrt x  + 8}}\\
 =  - 1 + \dfrac{{15}}{{\sqrt x  + 8}}\\
F \in Z\\
 \to \dfrac{{15}}{{\sqrt x  + 8}} \in Z\\
 \to \sqrt x  + 8 \in U\left( {15} \right)\\
 \to \sqrt x  + 8 = 15\\
Do:\sqrt x  + 8 \ge 8\forall x \ge 0\\
 \to \sqrt x  = 7\\
 \to x = 49\\
6)G = \dfrac{{3\sqrt x  - 2}}{{3\sqrt x  - 4}} = \dfrac{{3\sqrt x  - 4 + 2}}{{3\sqrt x  - 4}}\\
 = 1 + \dfrac{2}{{3\sqrt x  - 4}}\\
G \in Z\\
 \to \dfrac{2}{{3\sqrt x  - 4}} \in Z\\
 \to 3\sqrt x  - 4 \in U\left( 2 \right)\\
 \to \left[ \begin{array}{l}
3\sqrt x  - 4 = 2\\
3\sqrt x  - 4 =  - 2\\
3\sqrt x  - 4 = 1\\
3\sqrt x  - 4 =  - 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\sqrt x  = 2\\
\sqrt x  = \dfrac{2}{3}\left( l \right)\\
\sqrt x  = \dfrac{5}{3}\left( l \right)\\
\sqrt x  = 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 4\\
x = 1
\end{array} \right.\\
7)H = \dfrac{{3\sqrt x }}{{\sqrt x  - 1}} = \dfrac{{3\left( {\sqrt x  - 1} \right) + 3}}{{\sqrt x  - 1}}\\
 = 3 + \dfrac{3}{{\sqrt x  - 1}}\\
H \in Z\\
 \to \dfrac{3}{{\sqrt x  - 1}} \in Z\\
 \to \sqrt x  - 1 \in U\left( 3 \right)\\
 \to \left[ \begin{array}{l}
\sqrt x  - 1 = 3\\
\sqrt x  - 1 = 1\\
\sqrt x  - 1 =  - 1
\end{array} \right.\\
 \to \left[ \begin{array}{l}
\sqrt x  = 4\\
\sqrt x  = 2\\
\sqrt x  = 0
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 16\\
x = 4\\
x = 0
\end{array} \right.\\
8)I = \dfrac{{3 - x}}{{\sqrt x  + 3}} = \dfrac{{9 - x - 6}}{{\sqrt x  + 3}}\\
 = \dfrac{{\left( {\sqrt x  + 3} \right)\left( {\sqrt x  - 3} \right) - 6}}{{\sqrt x  + 3}}\\
 = \sqrt x  - 3 - \dfrac{6}{{\sqrt x  + 3}}\\
I \in Z\\
 \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{6}{{\sqrt x  + 3}} \in Z\\
\sqrt x  \in {Z^ + }
\end{array} \right.\\
 \Leftrightarrow \sqrt x  + 3 \in U\left( 6 \right)\\
 \to \left[ \begin{array}{l}
\sqrt x  + 3 = 6\\
\sqrt x  + 3 = 3
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 9\\
x = 0
\end{array} \right.\left( {TM} \right)
\end{array}\)