\(\begin{array}{l}
\quad M = \left(\dfrac{1}{\sqrt x + 3} + \dfrac{\sqrt x + 9}{x - 9}\right)\cdot \dfrac{\sqrt x}{2}\qquad (x\geqslant 0;x\ne 9)\\
1)\quad M = \left[\dfrac{1}{\sqrt x + 3} + \dfrac{\sqrt x + 9}{\left(\sqrt x - 3\right)\left(\sqrt x + 3\right)} \right]\cdot \dfrac{\sqrt x}{2}\\
\Leftrightarrow M = \dfrac{\sqrt x - 3 + \sqrt x + 9}{\left(\sqrt x - 3\right)\left(\sqrt x + 3\right)}\cdot \dfrac{\sqrt x}{2}\\
\Leftrightarrow M = \dfrac{2\sqrt x + 6}{\left(\sqrt x - 3\right)\left(\sqrt x + 3\right)}\cdot \dfrac{\sqrt x}{2}\\
\Leftrightarrow M = \dfrac{2\left(\sqrt x + 3\right)}{\left(\sqrt x - 3\right)\left(\sqrt x + 3\right)}\cdot \dfrac{\sqrt x}{2}\\
\Leftrightarrow M = \dfrac{\sqrt x}{\sqrt x-3}\\
b)\quad M > 1\\
\Leftrightarrow \dfrac{\sqrt x}{\sqrt x - 3} >1\\
\Leftrightarrow \dfrac{\sqrt x}{\sqrt x - 3} -1 >0\\
\Leftrightarrow \dfrac{\sqrt x - \left(\sqrt x - 3\right)}{\sqrt x - 3} >0\\
\Leftrightarrow \dfrac{3}{\sqrt x - 3} >0\\
\Leftrightarrow \sqrt x - 3 <0\\
\Leftrightarrow \sqrt x < 3\\
\Rightarrow x < 9\\
\text{Kết hợp ĐKXĐ ta được:}\\
0\leqslant x <9
\end{array}\)
