\(P=\dfrac{x+3}{\sqrt{x}+3}\) (ĐK: \(x\ge0\))
Mà: \(x\ge0\Rightarrow\left\{{}\begin{matrix}x+3\ge3\\\sqrt{x}+3\ge3\end{matrix}\right.\) nên:
\(P=\dfrac{x+3}{\sqrt{x}+3}\ge\dfrac{3}{3}=1\)
Dấu "=" xảy ra:
\(\dfrac{x+3}{\sqrt{x}+3}=1\)
\(\Leftrightarrow x=\sqrt{x}\)
\(\Leftrightarrow x=0\left(tm\right)\)
Vậy: \(P_{min}=1\) khi \(x=0\)