\(ĐK:x\ge0\\ Q=\dfrac{x-9+25}{\sqrt{x}+3}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+25}{\sqrt{x}+3}\\ Q=\sqrt{x}-3+\dfrac{25}{\sqrt{x}+3}\\ Q=\left(\sqrt{x}+3\right)+\dfrac{25}{\sqrt{x}+3}-6\ge2\sqrt{\left(\sqrt{x}+3\right)\cdot\dfrac{25}{\sqrt{x}+3}}-6\\ Q\ge2\sqrt{25}-6=10-6=4\\ Q_{min}=4\Leftrightarrow\sqrt{x}+3=5\left(\sqrt{x}+3>0\right)\Leftrightarrow x=4\left(tm\right)\)