\(B=\dfrac{2}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}+15}{x-9}\\ =\dfrac{2\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)+\sqrt{x}+15}{x-9}\\ =\dfrac{2\sqrt{x}-6+x+3\sqrt{x}+\sqrt{x}+15}{x-9}\\ =\dfrac{x+6\sqrt{x}+9}{\left(x-9\right)}\\ =\dfrac{\left(\sqrt{x}+3\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{\sqrt{x}+3}{\sqrt{x}-3}\)
\(B=\dfrac{2}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{\sqrt{x}+15}{x-9}\left(x\ge0;x\ne9\right)\)
\(=\dfrac{2\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)+\sqrt{x}+15}{x-9}\)
\(=\dfrac{2\sqrt{x}-6+x+3\sqrt{x}+\sqrt{x}+15}{x-9}\)
\(=\dfrac{x+6\sqrt{x}+9}{x-9}=\dfrac{\left(\sqrt{x}+3\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+3}{\sqrt{x}-3}\)
