Đáp án:
$\begin{array}{l}
a)M = A:B = \left( {\frac{3}{{\sqrt x - 1}} + \frac{{\sqrt x + 3}}{{x - 1}}} \right):\left( {\frac{{x + 2}}{{x + \sqrt x - 2}} - \frac{{\sqrt x }}{{\sqrt x + 2}}} \right)\\
= \frac{{3\left( {\sqrt x + 1} \right) + \sqrt x + 3}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}:\left[ {\frac{{x + 2}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}} - \frac{{\sqrt x }}{{\sqrt x + 2}}} \right]\\
= \frac{{4\sqrt x + 6}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}:\frac{{x + 2 - \sqrt x \left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}\\
= \frac{{4\sqrt x + 6}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}.\frac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 1} \right)}}{{\sqrt x + 2}}\\
= \frac{{4\sqrt x + 6}}{{\sqrt x + 1}}\\
b)x \ge 0;x \ne 1\\
M = \frac{{4\sqrt x + 6}}{{\sqrt x + 1}}\\
= \frac{{4\sqrt x + 4 + 2}}{{\sqrt x + 1}}\\
= 4 + \frac{2}{{\sqrt x + 1}}\\
Do:\sqrt x \ge 0\forall x \ge 0;x \ne 1\\
\Rightarrow \frac{1}{{\sqrt x + 1}} \le 1\forall x \ge 0;x \ne 1\\
\Rightarrow 4 + \frac{2}{{\sqrt x + 1}} \le 6\forall x \ge 0;x \ne 1\\
\Rightarrow M \le 6\forall x \ge 0;x \ne 1\\
\Rightarrow GTLN:M = 6 \Leftrightarrow x = 0
\end{array}$
