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\(C>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\x< -1\end{matrix}\right.\)

Vậy .......

\(a.\\ P=\dfrac{x-\sqrt{x}+\sqrt{x}-3-\sqrt{x}-3}{x-9}=\dfrac{x-\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)

\(b.\\ b.1.\\ x=\sqrt{4+4\sqrt{2}+2}+\sqrt{4-4\sqrt{2}+2}=\sqrt{\left(2+\sqrt{2}\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}=\left|2+\sqrt{2}\right|+\left|2-\sqrt{2}\right|=2+\sqrt{2}+2-\sqrt{2}=4\)

\(b.2.\\ x=\dfrac{\sqrt{2}+1-\sqrt{2}+1}{2-1}=2\)