\(M=\dfrac{1}{\sqrt{x}+3}+\dfrac{\sqrt{x}+9}{x-9}=\dfrac{1}{\sqrt{x}+3}+\dfrac{\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}-3+\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{2\sqrt{x}+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{2\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{2}{\sqrt{x}-3}\)
Để M là số tự nhiên \(\Rightarrow\left\{{}\begin{matrix}2⋮\sqrt{x}-3\\\sqrt{x}-3>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\sqrt{x}-3\in\left\{2;1;-1;-2\right\}\\x>9\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\in\left\{25;16;4;1\right\}\\x>9\end{matrix}\right.\Rightarrow x\in\left\{25;16\right\}\)
Thế vào M,ta đường \(\left\{{}\begin{matrix}x=25\Rightarrow M=1\\x=16\Rightarrow M=2\end{matrix}\right.\)
\(\Rightarrow M\) có giá trị là số tự nhiên lớn nhất là \(2\) khi \(x=16\)
