Đáp án:

$\begin{array}{l}
d)B = \dfrac{{\sqrt x }}{{\sqrt x  + 2}}\\
B < \dfrac{1}{2}\\
 \Leftrightarrow \dfrac{{\sqrt x }}{{\sqrt x  + 2}} - \dfrac{1}{2} < 0\\
 \Leftrightarrow \dfrac{{2\sqrt x  - \sqrt x  - 2}}{{2\left( {\sqrt x  + 2} \right)}} < 0\\
 \Leftrightarrow \sqrt x  - 2 < 0\\
 \Leftrightarrow \sqrt x  < 2\\
 \Leftrightarrow x < 4\\
 \Leftrightarrow 0 \le x < 4\\
{x_{max}} = 3\\
Vay\,x = 3\\
e)3B = 3.\dfrac{{\sqrt x }}{{\sqrt x  + 2}}\\
 = \dfrac{{3\sqrt x  + 6 - 6}}{{\sqrt x  + 2}}\\
 = 3 - \dfrac{6}{{\sqrt x  + 2}} \in Z\\
 \Leftrightarrow \dfrac{6}{{\sqrt x  + 2}} \in Z\\
 \Leftrightarrow \sqrt x  + 2 \in \left\{ {2;3;6} \right\}\\
 \Leftrightarrow \sqrt x  \in \left\{ {0;1;4} \right\}\\
 \Leftrightarrow x \in \left\{ {0;1;16} \right\}\left( {tmdk} \right)\\
Vay\,x \in \left\{ {0;1;16} \right\}\\
f)x \in \left\{ {0;1;16} \right\}\\
g)B = \dfrac{1}{m}\\
 \Leftrightarrow \dfrac{{\sqrt x }}{{\sqrt x  + 2}} = \dfrac{1}{m}\\
 \Leftrightarrow m\sqrt x  = \sqrt x  + 2\\
 \Leftrightarrow \left( {m - 1} \right).\sqrt x  = 2\\
 \Leftrightarrow \left\{ \begin{array}{l}
m \ne 1\\
\sqrt x  = \dfrac{2}{{m - 1}} \ge 0\\
\sqrt x  \ne 2
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
m \ne 1\\
m - 1 > 0\\
\dfrac{2}{{m - 1}} \ne 2
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m > 1\\
m - 1 \ne 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
m > 1\\
m \ne 2
\end{array} \right.\\
Vay\,m > 1;m \ne 2\\
h)B = \dfrac{{\sqrt x }}{{\sqrt x  + 2}} = \dfrac{{\sqrt x  + 2 - 2}}{{\sqrt x  + 2}}\\
 = 1 - \dfrac{2}{{\sqrt x  + 2}}\\
Do:\sqrt x  + 2 \ge 2\\
 \Leftrightarrow \dfrac{1}{{\sqrt x  + 2}} \le \dfrac{1}{2}\\
 \Leftrightarrow  - \dfrac{2}{{\sqrt x  + 2}} \ge  - 1\\
 \Leftrightarrow 1 - \dfrac{2}{{\sqrt x  + 2}} \ge 0\\
 \Leftrightarrow B \ge 0\\
 \Leftrightarrow GTNN:B = 0\,khi:x = 0
\end{array}$