Đáp án:

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

Giải thích các bước giải:

\(\begin{array}{l}
a.DK:x \ne  - 3\\
\dfrac{5}{{x + 3}} \in Z\\
 \Leftrightarrow x + 3 \in U\left( 5 \right)\\
 \to \left[ \begin{array}{l}
x + 3 = 5\\
x + 3 =  - 5\\
x + 3 = 1\\
x + 3 =  - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 2\\
x =  - 8\\
x =  - 2\\
x =  - 4
\end{array} \right.\\
b.DK:x \ne  - 3\\
B = \dfrac{{x + 5}}{{x + 3}} = \dfrac{{x + 3 + 2}}{{x + 3}} = 1 + \dfrac{2}{{x + 3}}\\
B \in Z\\
 \to \dfrac{2}{{x + 3}} \in Z\\
 \to x + 3 \in U\left( 2 \right)\\
 \to \left[ \begin{array}{l}
x + 3 = 2\\
x + 3 =  - 2\\
x + 3 = 1\\
x + 3 =  - 1
\end{array} \right. \to \left[ \begin{array}{l}
x =  - 1\\
x =  - 5\\
x =  - 2\\
x =  - 4
\end{array} \right.\\
c.DK:x \ne  - 3\\
C = \dfrac{{2x}}{{x + 3}} = \dfrac{{2\left( {x + 3} \right) - 6}}{{x + 3}}\\
 = 2 - \dfrac{6}{{x + 3}}\\
C \in Z\\
 \Leftrightarrow \dfrac{6}{{x + 3}} \in Z\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 3 = 6\\
x + 3 =  - 6\\
x + 3 = 3\\
x + 3 =  - 3\\
x + 3 = 2\\
x + 3 =  - 2\\
x + 3 = 1\\
x + 3 =  - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 3\\
x =  - 9\\
x = 0\\
x =  - 6\\
x =  - 1\\
x =  - 5\\
x =  - 2\\
x =  - 4
\end{array} \right.\\
d.D = \dfrac{{x - 2}}{{x + 3}} = \dfrac{{x + 3 - 5}}{{x + 3}} = 1 - \dfrac{5}{{x + 3}}\\
D \in Z\\
 \Leftrightarrow \dfrac{5}{{x + 3}} \in Z\\
 \Leftrightarrow x + 3 \in U\left( 5 \right)\\
 \to \left[ \begin{array}{l}
x + 3 = 5\\
x + 3 =  - 5\\
x + 3 = 1\\
x + 3 =  - 1
\end{array} \right. \to \left[ \begin{array}{l}
x = 2\\
x =  - 8\\
x =  - 2\\
x =  - 4
\end{array} \right.\\
f.DK:x \ge 0\\
\dfrac{5}{{\sqrt x  + 1}} \in Z\\
 \to \sqrt x  + 1 \in U\left( 5 \right)\\
 \to \left[ \begin{array}{l}
\sqrt x  + 1 = 5\\
\sqrt x  + 1 =  - 5\\
\sqrt x  + 1 = 1\\
\sqrt x  + 1 =  - 1
\end{array} \right. \to \left[ \begin{array}{l}
\sqrt x  = 4\\
\sqrt x  =  - 6\left( l \right)\\
\sqrt x  = 0\\
\sqrt x  =  - 2\left( l \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 16\\
x = 0
\end{array} \right.\\
g.G = \dfrac{{\sqrt x  + 3}}{{\sqrt x  + 1}} = \dfrac{{\sqrt x  + 1 + 2}}{{\sqrt x  + 1}} = 1 + \dfrac{2}{{\sqrt x  + 1}}\\
G \in Z\\
 \to \dfrac{2}{{\sqrt x  + 1}} \in Z\\
 \to \sqrt x  + 1 \in U\left( 2 \right)\\
 \to \left[ \begin{array}{l}
\sqrt x  + 1 = 2\\
\sqrt x  + 1 =  - 2\\
\sqrt x  + 1 = 1\\
\sqrt x  + 1 =  - 1
\end{array} \right. \to \left[ \begin{array}{l}
\sqrt x  = 1\\
\sqrt x  =  - 3\left( l \right)\\
\sqrt x  = 0\\
\sqrt x  =  - 2\left( l \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
x = 1\\
x = 0
\end{array} \right.\\
h.H = \dfrac{{\sqrt x  - 2}}{{\sqrt x  + 1}} = \dfrac{{\sqrt x  + 1 - 3}}{{\sqrt x  + 1}} = 1 - \dfrac{3}{{\sqrt x  + 1}}\\
H \in Z\\
 \Leftrightarrow \dfrac{3}{{\sqrt x  + 1}} \in Z\\
 \Leftrightarrow \sqrt x  + 1 \in U\left( 3 \right)\\
 \to \left[ \begin{array}{l}
\sqrt x  + 1 = 3\\
\sqrt x  + 1 =  - 3\left( l \right)\\
\sqrt x  + 1 = 1\\
\sqrt x  + 1 =  - 1\left( l \right)
\end{array} \right. \to \left[ \begin{array}{l}
x = 4\\
x = 0
\end{array} \right.
\end{array}\)