Ta biết: \(\sqrt{P}=\dfrac{1}{2}\Rightarrow P=\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}\) (1)
Với đk: \(P\ge0\)
\(\Rightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\ge0\)
\(\Leftrightarrow\sqrt{x}-2\ge0\) (vì \(\sqrt{x}+1\ge1>0\forall x\ge0\))
\(\Leftrightarrow\sqrt{x}\ge2\)
\(\Leftrightarrow x\ge4\)
\(\left(1\right)\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{1}{4}\)
\(\Leftrightarrow4\left(\sqrt{x}-2\right)=\sqrt{x}+1\)
\(\Leftrightarrow4\sqrt{x}-8=\sqrt{x}+1\)
\(\Leftrightarrow4\sqrt{x}-\sqrt{x}=1+8\)
\(\Leftrightarrow3\sqrt{x}=9\)
\(\Leftrightarrow\sqrt{x}=3\)
\(\Leftrightarrow x=3^2\)
\(\Leftrightarrow x=9\left(tm\right)\)
Vậy: ...
