a: ĐKXĐ: x<>9; x>=0
\(A=\dfrac{\sqrt{x}+3-\sqrt{x}+3}{\left(\sqrt{x}-3\right)\cdot\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{3}=\dfrac{2}{\sqrt{x}+3}\)
b: Để A>1/3 thì A-1/3>0
\(\Leftrightarrow\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{3}>0\)
\(\Leftrightarrow6-\sqrt{x}-3>0\)
\(\Leftrightarrow3-\sqrt{x}>0\)
=>0<x<9