Question

\(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

Answer 1

\(P+\sqrt{x}\ge0\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\sqrt{x}\ge0\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1+\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\ge0\)

\(\Leftrightarrow\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-3}\ge0\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-3}\ge0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}-1=0\\\sqrt{x}-3>0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x>9\end{matrix}\right.\)