\[\begin{array}{l}
M = \left( {\frac{{\sqrt x }}{{x - 4}} + \frac{1}{{\sqrt x - 2}}} \right):\frac{2}{{\sqrt x - 2}}\\
dk:x \ge 0;x \ne 4\\
= \left( {\frac{{\sqrt x }}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}} + \frac{{\sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}} \right).\frac{{\sqrt x - 2}}{2}\\
= \frac{{\sqrt x + \sqrt x + 2}}{{\left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}.\frac{{\sqrt x - 2}}{2}\\
= \frac{{2\sqrt x + 2}}{{\sqrt x + 2}}.2\\
= \frac{{\sqrt x + 1}}{{\sqrt x + 2}}\\
Xet:{M^2} - M = {\left( {\frac{{\sqrt x + 1}}{{\sqrt x + 2}}} \right)^2} - \frac{{\sqrt x + 1}}{{\sqrt x + 2}}\\
= \frac{{{{\left( {\sqrt x + 1} \right)}^2} - \left( {\sqrt x + 1} \right)\left( {\sqrt x + 2} \right)}}{{{{\left( {\sqrt x + 2} \right)}^2}}}\\
= \frac{{x + 2\sqrt x + 1 - x - 2\sqrt x - \sqrt x - 2}}{{{{\left( {\sqrt x + 2} \right)}^2}}}\\
= \frac{{ - \sqrt x - 1}}{{{{\left( {\sqrt x + 2} \right)}^2}}} < 0\forall x \ge 0;x \ne 4\\
= > {M^2} - M < 0\\
= > {M^2} < M\\
Vay:{M^2} < M
\end{array}\]

.

$Vậy:M^2<M$.