\(ĐKXĐ:x\ge0;x\ne4\)
\(P=\dfrac{2\sqrt{x}-2}{\sqrt{x}+2}=\dfrac{2\left(\sqrt{x}+2\right)-6}{\sqrt{x}+2}=2-\dfrac{6}{\sqrt{x}+2}\)
Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+2\ge2\Rightarrow\dfrac{6}{\sqrt{x}+2}\le3\)
Suy ra: \(P\ge2-3=-1\)
Dấu "=" xảy ra \(\Leftrightarrow x=0\)
Vậy \(MinP=-1\)