\(a)P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}+\dfrac{6\sqrt{x}-4}{1-x}\\
P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\\
P=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1}+\dfrac{3\left(\sqrt{x}-1\right)}{x-1}-\dfrac{6\sqrt{x}-4}{x-1}\\
P=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{x-1}\\
P=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\
P=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\
P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\\
b)P=-1\\
\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=-1\\
\Leftrightarrow\sqrt{x}-1=-\sqrt{x}-1\\
\Leftrightarrow\sqrt{x}+\sqrt{x}=-1+1\\
\Leftrightarrow2\sqrt{x}=0\\
\Leftrightarrow\sqrt{x}=0\\
\Leftrightarrow x=0\left(TMĐKXĐ\right)\)
Vậy \(x=0\) thì \(P=-1\)
\(c)\) Xét hiệu
\(P-1\\
=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-1\\
=\dfrac{\sqrt{x}-1-\sqrt{x}-1}{\sqrt{x}+1}\\
=\dfrac{-2}{\sqrt{x}+1}\)
Mà \(\left\{{}\begin{matrix}-2< 0\\x\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}-2< 0\\\sqrt{x}\ge0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}-2< 0\\\sqrt{x}+1\ge1\end{matrix}\right.\)
\(\Rightarrow\dfrac{-2}{\sqrt{x}+1}\le-2\\
\Rightarrow P-1\le-2\\
\Rightarrow P\le-1\)
