\(A=\dfrac{x-9+9}{\sqrt{x}-3}=\sqrt{x}+3+\dfrac{9}{\sqrt{x}-3}\)
\(=\sqrt{x}-3+\dfrac{9}{\sqrt{x}-3}+6\ge2\cdot3+6=12\)
Dấu '=' xảy ra khi \(\left[{}\begin{matrix}\sqrt{x}-3=-3\\\sqrt{x}-3=3\end{matrix}\right.\Leftrightarrow x\in\left\{0;36\right\}\)