a.
\(x=9\Rightarrow A=\dfrac{\sqrt{9}+1}{\sqrt{9}}=\dfrac{3+1}{3}=\dfrac{4}{3}\)
b.
\(B=\dfrac{x-3\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-3\sqrt{x}+4-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{x-4\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
c.
\(P=\dfrac{B}{A}=\dfrac{\sqrt{x}-2}{\sqrt{x}}:\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
\(P+2=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+2=\dfrac{3\sqrt{x}}{\sqrt{x}+1}\)
Do \(\left\{{}\begin{matrix}\sqrt{x}>0\\\sqrt{x}+1>0\end{matrix}\right.\) ; \(\forall x>0\)
\(\Rightarrow\dfrac{3\sqrt{x}}{\sqrt{x}+1}>0\Rightarrow P+2>0\)
\(\Rightarrow P>-2\)
