\(=\dfrac{x-9+16}{\sqrt{x}+3}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+16}{\sqrt{x}+3}\\ =\sqrt{x}-3+\dfrac{16}{\sqrt{x}+3}=\sqrt{x}+3+\dfrac{16}{\sqrt{x}+3}-6\left(1\right)\)
Áp dụng BĐT cosi: \(\left(1\right)\ge2\sqrt{16}-6=2\)
Dấu \("="\Leftrightarrow\left(\sqrt{x}+3\right)^2=16\Leftrightarrow\sqrt{x}+3=4\Leftrightarrow x=1\left(tm\right)\)
Vậy GTNN là 2, xảy ra khi x=1