Ta có: \(\sqrt{P}< \dfrac{1}{2}\Rightarrow P< \left(\dfrac{1}{2}\right)^2\Leftrightarrow P< \dfrac{1}{4}\) (1)
Với đk: \(P\ge0\)
\(\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\ge0\)
\(\Leftrightarrow\sqrt{x}-2\ge0\) (vì \(\sqrt{x}+1>0\forall x\ge0\))
\(\Leftrightarrow\sqrt{x}\ge2\)
\(\Leftrightarrow x\ge4\)
\(\left(1\right)\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< \dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}-\dfrac{1}{4}< 0\)
\(\Leftrightarrow\dfrac{4\sqrt{x}-8}{4\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+1}{4\left(\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\dfrac{4\sqrt{x}-8-\sqrt{x}-1}{4\left(\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow\dfrac{3\sqrt{x}-9}{4\left(\sqrt{x}+1\right)}< 0\)
\(\Leftrightarrow3\sqrt{x}-9< 0\)
\(\Leftrightarrow\sqrt{x}< 3\)
\(\Leftrightarrow x< 9\)
Kết hợp với đk: \(4\le x< 9\)
