ĐKXĐ: \(x\ge0\)
\(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}\)
Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\Rightarrow\dfrac{2}{\sqrt{x}+1}\le2\Rightarrow1-\dfrac{2}{\sqrt{x}+1}\ge-1\)
\(\Rightarrow P_{min}=-1\) khi \(x=0\)