$\begin{array}{l} A = \sqrt x + \dfrac{{36}}{{\sqrt x - 3}}\\ A = \sqrt x - 3 + \dfrac{{36}}{{\sqrt x - 3}} + 3 \ge 2\sqrt {\left( {\sqrt x - 3} \right).\dfrac{{36}}{{\sqrt x - 3}}} + 3 = 2.\sqrt {36} + 3 = 15 \end{array}$
Dấu bằng xảy ra khi và chỉ khi
$\begin{array}{l} \sqrt x - 3 = \dfrac{{36}}{{\sqrt x - 3}}\\ \Leftrightarrow {\left( {\sqrt x - 3} \right)^2} = 36\\ \Leftrightarrow \sqrt x - 3 = 6\left( {do \,x > 9\Rightarrow \sqrt x-3>0} \right)\\ \Leftrightarrow \sqrt x = 9\\ \Leftrightarrow x = 81 \end{array}$