ĐKXĐ: \(x\ge0\)
\(A=\dfrac{3\left(\sqrt{x}+2\right)}{\sqrt{x}+2}-\dfrac{6}{\sqrt{x}+2}=3-\dfrac{6}{\sqrt{x}+2}\)
Do \(\sqrt{x}+2\ge2\Leftrightarrow\dfrac{6}{\sqrt{x}+2}\le\dfrac{6}{2}=3\)
\(\Leftrightarrow-\dfrac{6}{\sqrt{x}+2}\ge-3\)
\(A=3-\dfrac{6}{\sqrt{x}+2}\ge3-3=0\)
\(minA=0\Leftrightarrow x=0\)