\(\begin{array}{l}
M = \frac{{\sqrt x + 6}}{{\sqrt x + 1}} = 1 + \frac{5}{{\sqrt x + 1}}\\
Do\,\,\sqrt x \ge 0 \Rightarrow \sqrt x + 1 > 0 \Rightarrow M > 1\\
Do\,\,\sqrt x + 1 \ge 1 \Leftrightarrow \frac{5}{{\sqrt x + 1}} \le 5 \Rightarrow M \le 6\\
Suy\,ra\,1 < M \le 6 \Rightarrow M \in \left\{ {2;3;4;5;6} \right\}\\
+ )M = 2 \Rightarrow 1 + \frac{5}{{\sqrt x + 1}} = 2 \Leftrightarrow \frac{5}{{\sqrt x + 1}} = 1 \Leftrightarrow x = 16\left( {tm} \right)\\
+ )M = 3 \Rightarrow 1 + \frac{5}{{\sqrt x + 1}} = 3 \Leftrightarrow \frac{5}{{\sqrt x + 1}} = 2 \Leftrightarrow x = \frac{9}{4}\left( {tm} \right)\\
+ )M = 4 \Rightarrow 1 + \frac{5}{{\sqrt x + 1}} = 4 \Leftrightarrow \frac{5}{{\sqrt x + 1}} = 3 \Leftrightarrow x = \frac{4}{9}\left( {tm} \right)\\
+ )M = 5 \Rightarrow 1 + \frac{5}{{\sqrt x + 1}} = 5 \Leftrightarrow \frac{5}{{\sqrt x + 1}} = 4 \Leftrightarrow x = \frac{1}{{16}}\left( {tm} \right)\\
+ )M = 6 \Rightarrow 1 + \frac{5}{{\sqrt x + 1}} = 6 \Leftrightarrow \frac{5}{{\sqrt x + 1}} = 5 \Leftrightarrow x = 0\left( {tm} \right)
\end{array}\)
