TXĐ: \(x\ge0\)
\(P=\dfrac{x+7}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{16}{\sqrt{x}+3}=\sqrt{x}+3+\dfrac{16}{\sqrt{x}+3}-6\ge2\sqrt{\dfrac{16\left(\sqrt{x}+3\right)}{\sqrt{x}+3}}-6=8-6=2\)
\(\Rightarrow P_{min}=2\)
Dấu "=" xảy ra khi \(\sqrt{x}+3=\dfrac{16}{\sqrt{x}+3}\Rightarrow\sqrt{x}+3=4\Rightarrow x=1\)