ĐKXĐ: \(x\ge0;x\ne4\)
Để \(P=\left|P\right|\Leftrightarrow P\ge0\)
\(\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}\ge0\Rightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2>0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x>4\end{matrix}\right.\)
ĐKXĐ: \(x\ge0;x\ne4\)
Để \(\left|P\right|=P\)
\(\Rightarrow P\ge0\\ \Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}\ge0\)
mà \(\sqrt{x}\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-2>0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\\sqrt{x}>2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x>4\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=0\\x>4\end{matrix}\right.\)thì |P| = P
