Đáp án:
$\begin{array}{l}
DK:x > 0\\
M = \dfrac{{x + 4}}{{\sqrt x }} + \dfrac{{x\sqrt x - 8}}{{x - 2\sqrt x }} - \dfrac{{x\sqrt x + 8}}{{x + 2\sqrt x }}\\
= \dfrac{{x + 4}}{{\sqrt x }} + \dfrac{{{{\left( {\sqrt x } \right)}^3} - 8}}{{\sqrt x \left( {\sqrt x - 2} \right)}} - \dfrac{{{{\left( {\sqrt x } \right)}^3} + 8}}{{\sqrt x \left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x + 4}}{{\sqrt x }} + \dfrac{{\left( {\sqrt x - 2} \right)\left( {x + 2\sqrt x + 4} \right)}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
- \dfrac{{\left( {\sqrt x + 2} \right)\left( {x - 2\sqrt x + 4} \right)}}{{\sqrt x \left( {\sqrt x + 2} \right)}}\\
= \dfrac{{x + 4}}{{\sqrt x }} + \dfrac{{x + 2\sqrt x + 4}}{{\sqrt x }} - \dfrac{{x - 2\sqrt x + 4}}{{\sqrt x }}\\
= \dfrac{{x + 4 + x + 2\sqrt x + 4 - \left( {x - 2\sqrt x + 4} \right)}}{{\sqrt x }}\\
= \dfrac{{x + 4\sqrt x + 4}}{{\sqrt x }}\\
= \sqrt x + 4 + \dfrac{4}{{\sqrt x }}\\
Theo\,Co - si:\sqrt x + \dfrac{4}{{\sqrt x }} \ge 2\sqrt {\sqrt x .\dfrac{4}{{\sqrt x }}} = 2.2 = 4\\
\Rightarrow \sqrt x + 4 + \dfrac{4}{{\sqrt x }} \ge 4 + 4 = 8\\
\Rightarrow M \ge 8\\
\Rightarrow GTNN:M = 8 > 4
\end{array}$
