a) Ta có: \(\sqrt{x}+2\ge2\)
\(\Rightarrow\dfrac{1}{\sqrt{x}+2}\le\dfrac{1}{2}\)\(\Rightarrow P=\dfrac{-1}{\sqrt{x}+2}\ge-\dfrac{1}{2}\)
\(minP=-\dfrac{1}{2}\Leftrightarrow x=0\)
b) Ta có: \(2\sqrt{x}+3\ge3\)
\(\Rightarrow\dfrac{5}{2\sqrt{x}+3}\le\dfrac{5}{3}\)\(\Rightarrow B=\dfrac{-5}{2\sqrt{x}+3}\ge-\dfrac{5}{3}\)
\(minB=-\dfrac{5}{3}\Leftrightarrow x=0\)
c) \(C=\dfrac{5\sqrt{x}+13}{\sqrt{x}+3}=\dfrac{5\left(\sqrt{x}+3\right)}{\sqrt{x}+3}-\dfrac{2}{\sqrt{x}+3}=5-\dfrac{2}{\sqrt{x}+3}\)
Do \(\sqrt{x}+3\ge3\Rightarrow\dfrac{2}{\sqrt{x}+3}\le\dfrac{2}{3}\)
\(\Rightarrow C=5-\dfrac{2}{\sqrt{x}+3}\ge5+-\dfrac{2}{3}=\dfrac{13}{3}\)
\(minC=\dfrac{13}{3}\Leftrightarrow x=0\)
