Đáp án:
\(\dfrac{{3\sqrt x }}{{\sqrt x + 1}}\)
Giải thích các bước giải:
\(\begin{array}{l}
DK:x \ge 0;x \ne 16\\
B = \dfrac{{2\left( {x + 4} \right)}}{{x - 3\sqrt x - 4}} + \dfrac{{\sqrt x }}{{\sqrt x + 1}} - \dfrac{8}{{\sqrt x - 4}}\\
= \dfrac{{2x + 8 + \sqrt x \left( {\sqrt x - 4} \right) - 8\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 4} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{2x + 8 + x - 4\sqrt x - 8\sqrt x - 8}}{{\left( {\sqrt x - 4} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{3x - 12\sqrt x }}{{\left( {\sqrt x - 4} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{3\sqrt x \left( {\sqrt x - 4} \right)}}{{\left( {\sqrt x - 4} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{3\sqrt x }}{{\sqrt x + 1}}
\end{array}\)
