a, ĐK: \(x\ge0,x\ne9\)
b, \(A=\left(\dfrac{1}{\sqrt{x}-3}-\dfrac{1}{\sqrt{x}+3}\right):\dfrac{3}{\sqrt{x}-3}\)
\(=\dfrac{\sqrt{x}+3-\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{3}{\sqrt{x}-3}\)
\(=\dfrac{6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{3}{\sqrt{x}-3}\)
\(=\dfrac{2}{\sqrt{x}+3}\)
c, ĐK: \(x\ge0,x\ne9\)
\(A>\dfrac{1}{3}\Leftrightarrow\dfrac{2}{\sqrt{x}+3}>\dfrac{1}{3}\Leftrightarrow\sqrt{x}+3>6\Leftrightarrow\sqrt{x}>3\Leftrightarrow x>9\)
Vậy \(A>\dfrac{1}{3}\Leftrightarrow x>9\)
d, ĐK: \(x\ge0,x\ne9\)
Ta có: \(x\ge0\forall x\Leftrightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+3\ge3\)\(\Leftrightarrow\dfrac{1}{\sqrt{x}+3}\le\dfrac{1}{3}\Leftrightarrow\dfrac{2}{\sqrt{x}+3}\le\dfrac{2}{3}\)
Dấu bằng xảy ra \(\Leftrightarrow\sqrt{x}+3=3\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\left(TM\right)\)
Vậy MaxA = \(\dfrac{2}{3}\Leftrightarrow x=0\)