Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
x \ge 0\\
x \ne 4
\end{array} \right.\\
M = \left( {\dfrac{1}{{\sqrt x + 2}} + \dfrac{7}{{x - 4}}} \right):\left( {\dfrac{{\sqrt x - 1}}{{\sqrt x - 2}} - 1} \right)\\
= \dfrac{{\sqrt x - 2 + 7}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}:\dfrac{{\sqrt x - 1 - \sqrt x + 2}}{{\sqrt x - 2}}\\
= \dfrac{{\sqrt x + 5}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right)}}.\dfrac{{\sqrt x - 2}}{1}\\
= \dfrac{{\sqrt x + 5}}{{\sqrt x + 2}}\\
b)x = 7 + 4\sqrt 3 \left( {tmdk} \right)\\
= 4 + 2.2.\sqrt 3 + 3\\
= {\left( {2 + \sqrt 3 } \right)^2}\\
\Rightarrow \sqrt x = 2 + \sqrt 3 \\
\Rightarrow M = \dfrac{{\sqrt x + 5}}{{\sqrt x + 2}} = \dfrac{{2 + \sqrt 3 + 5}}{{2 + \sqrt 3 + 2}}\\
= \dfrac{{7 + \sqrt 3 }}{{4 + \sqrt 3 }} = \dfrac{{25 - 3\sqrt 3 }}{{13}}\\
c)M < 2\\
\Rightarrow \dfrac{{\sqrt x + 5}}{{\sqrt x + 2}} < 2\\
\Rightarrow \sqrt x + 5 < 2\left( {\sqrt x + 2} \right)\\
\Rightarrow \sqrt x + 5 < 2\sqrt x + 4\\
\Rightarrow \sqrt x > 1\\
\Rightarrow x > 1\\
Vay\,x > 1\,khi:M < 2
\end{array}$
