Đáp án:
$\begin{array}{l}
1)Dkxd:x \ge 0;x \ne 16\\
B = \dfrac{{2\left( {x + 4} \right)}}{{x - 3\sqrt x - 4}} + \dfrac{{\sqrt x }}{{\sqrt x + 1}} - \dfrac{8}{{\sqrt x - 4}}\\
= \dfrac{{2x + 8 + \sqrt x \left( {\sqrt x - 4} \right) - 8\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 4} \right)}}\\
= \dfrac{{2x + 8 + x - 4\sqrt x - 8\sqrt x - 8}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 4} \right)}}\\
= \dfrac{{3x - 12\sqrt x }}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 4} \right)}}\\
= \dfrac{{3\sqrt x \left( {\sqrt x - 4} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 4} \right)}}\\
= \dfrac{{3\sqrt x }}{{\sqrt x + 1}}\\
2)B \le \dfrac{3}{2}\\
\Rightarrow \dfrac{{3\sqrt x }}{{\sqrt x + 1}} \le \dfrac{3}{2}\\
\Rightarrow \dfrac{{\sqrt x }}{{\sqrt x + 1}} \le \dfrac{1}{2}\\
\Rightarrow \dfrac{{2\sqrt x - \sqrt x - 1}}{{2\left( {\sqrt x + 1} \right)}} \le 0\\
\Rightarrow \sqrt x - 1 \le 0\\
\Rightarrow \sqrt x \le 1\\
\Rightarrow x \le 1\\
Vay\,0 \le x \le 1\\
3)\sqrt {2x - 1} = x\\
\Rightarrow 2x - 1 = {x^2}\\
\Rightarrow {x^2} - 2x + 1 = 0\\
\Rightarrow {\left( {x - 1} \right)^2} = 0\\
\Rightarrow x = 1\left( {tmdk} \right)\\
\Rightarrow B = \dfrac{{3\sqrt x }}{{\sqrt x + 1}} = \dfrac{3}{2}\\
4)B = \dfrac{{3\sqrt x }}{{\sqrt x + 1}} = \dfrac{{3\sqrt x + 3 - 3}}{{\sqrt x + 1}}\\
= 3 - \dfrac{3}{{\sqrt x + 1}} \in Z\\
\Rightarrow \dfrac{3}{{\sqrt x + 1}} \in Z\\
\Rightarrow \left( {\sqrt x + 1} \right) \in \left\{ {1;3} \right\}\\
\Rightarrow \left[ \begin{array}{l}
\sqrt x = 0\\
\sqrt x = 2
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 0\left( {tmdk} \right)\\
x = 4\left( {tmdk} \right)
\end{array} \right.
\end{array}$
