Đáp án:
$\begin{array}{l}
a)A = \left( {\frac{{x + 5\sqrt x - 27}}{{x - 16}} + \frac{{3 - \sqrt x }}{{\sqrt x - 4}}} \right):\frac{1}{{\sqrt x + 4}}\\
= \frac{{x + 5\sqrt x - 27 + \left( {3 - \sqrt x } \right)\left( {\sqrt x + 4} \right)}}{{\left( {\sqrt x - 4} \right)\left( {\sqrt x + 4} \right)}}.\left( {\sqrt x + 4} \right)\\
= \frac{{x + 5\sqrt x - 27 - x - 4\sqrt x + 3\sqrt x + 12}}{{\sqrt x - 4}}\\
= \frac{{4\sqrt x - 15}}{{\sqrt x - 4}}\\
b)x \ge 0;x \ne 16\\
B = - 2A\\
\Rightarrow \sqrt x - 4 = - 2.\frac{{4\sqrt x - 15}}{{\sqrt x - 4}}\\
\Rightarrow x - 8\sqrt x + 16 = - 8\sqrt x + 30\\
\Rightarrow x = 14\left( {tm} \right)\\
c)M = A:B\\
= \frac{{4\sqrt x - 15}}{{\sqrt x - 4}}:\left( {\sqrt x - 4} \right)\\
= \frac{{4\sqrt x - 15}}{{{{\left( {\sqrt x - 4} \right)}^2}}}\\
= \frac{{4\left( {\sqrt x - 4} \right) + 1}}{{{{\left( {\sqrt x - 4} \right)}^2}}}\\
= \frac{4}{{\sqrt x - 4}} + \frac{1}{{{{\left( {\sqrt x - 4} \right)}^2}}}\\
= {\left( {\frac{1}{{\sqrt x - 4}}} \right)^2} + 2.2.\frac{1}{{\sqrt x - 4}} + 4 - 4\\
= {\left( {\frac{1}{{\sqrt x - 4}} + 2} \right)^2} - 4 \ge - 4\forall x \ge 0;x \ne 16\\
\Rightarrow GTNN:M = - 4 \Leftrightarrow \frac{1}{{\sqrt x - 4}} = - 2\\
\Rightarrow \sqrt x = - \frac{1}{2} + 4 = \frac{7}{2}\\
\Rightarrow x = \frac{{49}}{4}\left( {tmdk} \right)
\end{array}$
