\[\begin{array}{l}
M = \left( {\frac{{\sqrt x }}{{\sqrt x - 2}} - \frac{{\sqrt x }}{{\sqrt x + 2}}} \right).\left( {1 - \frac{4}{x}} \right)\\
a)dk:x > 0;x \ne 4\\
= \frac{{\sqrt x \left( {\sqrt x + 2} \right) - \sqrt x \left( {\sqrt x - 2} \right)}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\frac{{x - 4}}{x}\\
= \frac{{x + 2\sqrt x - x + 2\sqrt x }}{{x - 4}}.\frac{{x - 4}}{x}\\
= \frac{{4\sqrt x }}{x}\\
vay:M = \frac{{4\sqrt x }}{x}\left( {x > 0;x \ne 4} \right)\\
b)M \ge 2 \Leftrightarrow \frac{{4\sqrt x }}{x} \ge 2\left( {x > 0;x \ne 4} \right)\\
\Leftrightarrow \frac{{4\sqrt x }}{x} - 2 \ge 0\\
\Leftrightarrow \frac{{4\sqrt x - 2{\rm{x}}}}{x} \ge 0\\
\Leftrightarrow 4\sqrt x - 2{\rm{x}} \ge 0\left( {do:x > 0} \right)\\
\Leftrightarrow 2\sqrt x \left( {2 - \sqrt x } \right) \ge 0\\
\Leftrightarrow 2 - \sqrt x \ge 0\left( {do:2\sqrt x > 0} \right)\\
\Leftrightarrow \sqrt x \le 2\\
\Leftrightarrow x \le 4
\end{array}\]

Kết hơph dkxd ta được 0<x<4 thì $M\ge 2$