a.
\(x=9\Rightarrow A=\dfrac{\sqrt{9}}{\sqrt{9}+2}=\dfrac{3}{3+2}=\dfrac{3}{5}\)
b.
\(B=\dfrac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
c.
\(\dfrac{A}{B}=\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}=-1+\dfrac{2\sqrt{x}}{\sqrt{x}+2}\)
Do \(\dfrac{2\sqrt{x}}{\sqrt{x}+2}>0;\forall x>0\)
\(\Rightarrow-1+\dfrac{2\sqrt{x}}{\sqrt{x}+2}>-1\)
