a: \(B=\dfrac{x+3+\sqrt{x}-3}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}=\dfrac{x+\sqrt{x}}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
b: \(B-\dfrac{1}{3}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}-\dfrac{1}{3}=\dfrac{3\sqrt{x}+3-\sqrt{x}-3}{\sqrt{x}+3}>0\)
=>B>1/3