Đáp án:
$\begin{array}{l}
a)x > 0;x \ne 1;x \ne 4\\
M = \left( {1 - \frac{{4\sqrt x }}{{x - 1}} + \frac{1}{{\sqrt x - 1}}} \right):\frac{{x - 2\sqrt x }}{{x - 1}}\\
= \left( {1 - \frac{{4\sqrt x }}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} + \frac{1}{{\sqrt x - 1}}} \right).\frac{{x - 1}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \frac{{x - 1 - 4\sqrt x + \sqrt x + 1}}{{x - 1}}.\frac{{x - 1}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \frac{{x - 3\sqrt x }}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \frac{{\sqrt x \left( {\sqrt x - 3} \right)}}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \frac{{\sqrt x - 3}}{{\sqrt x - 2}}\\
b)M = \frac{1}{2}\\
\Rightarrow \frac{{\sqrt x - 3}}{{\sqrt x - 2}} = \frac{1}{2}\\
\Rightarrow 2\sqrt x - 6 = \sqrt x - 2\\
\Rightarrow \sqrt x = 4\\
\Rightarrow x = 16\left( {tmdk} \right)\\
Vậy\,x = 16
\end{array}$
