\(1,B=\dfrac{1}{2-3}=\dfrac{1}{-1}=-1\\ 2,S=A-B=\dfrac{x+3}{x-9}+\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-3}\\ S=\dfrac{x+3+2\sqrt{x}-6-\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{x+\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ S=\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\\ 3,S=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}-3}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(\Leftrightarrow x-4=x-2\sqrt{x}-3\\ \Leftrightarrow2\sqrt{x}=1\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)
