Đáp án:
$\begin{array}{l}
x \ge 0;x \ne 1;x \ne 9\\
a)x = 4 - 2\sqrt 3 \left( {tmdk} \right)\\
= {\left( {\sqrt 3 - 1} \right)^2}\\
\Leftrightarrow \sqrt x = \sqrt 3 - 1\\
B = \dfrac{5}{{\sqrt x - 3}} = \dfrac{5}{{\sqrt 3 - 1 - 3}} = \dfrac{5}{{\sqrt 3 - 4}}\\
= \dfrac{{5\left( {\sqrt 3 + 4} \right)}}{{3 - {4^2}}}\\
= \dfrac{{ - 5\sqrt 3 - 20}}{{13}}\\
b)M = B:A\\
= \dfrac{5}{{\sqrt x - 3}}:\left( {\dfrac{{4\sqrt x }}{{x - 1}} + \dfrac{3}{{\sqrt x + 1}} - \dfrac{2}{{\sqrt x - 1}}} \right)\\
= \dfrac{5}{{\sqrt x - 3}}:\dfrac{{4\sqrt x + 3\left( {\sqrt x - 1} \right) - 2\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}\\
= \dfrac{5}{{\sqrt x - 3}}.\dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}{{4\sqrt x + 3\sqrt x - 3 - 2\sqrt x - 2}}\\
= \dfrac{5}{{\sqrt x - 3}}.\dfrac{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}{{5\sqrt x - 5}}\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}}\\
c)M < \dfrac{2}{3}\\
\Leftrightarrow \dfrac{{\sqrt x + 1}}{{\sqrt x - 3}} - \dfrac{2}{3} < 0\\
\Leftrightarrow \dfrac{{3\sqrt x + 3 - 2\sqrt x + 6}}{{3\left( {\sqrt x - 3} \right)}} < 0\\
\Leftrightarrow \dfrac{{\sqrt x + 9}}{{3\left( {\sqrt x - 3} \right)}} < 0\\
\Leftrightarrow \sqrt x - 3 < 0\\
\Leftrightarrow \sqrt x < 3\\
\Leftrightarrow x < 9\\
Vậy\,0 \le x < 9;x \ne 1
\end{array}$
