ĐK: $x>0, x\ne 1$
a,
$A=\Big(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\Big):\dfrac{1}{3\sqrt{x}}(x-1)$
$=\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}(\sqrt{x}-1)}.3\sqrt{x}(x-1)$
$=\dfrac{3\sqrt{x}(\sqrt{x}-1)(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}-1)}$
$=3(\sqrt{x}+1)$
b,
$A=9\Rightarrow 3(\sqrt{x}+1)=9$
$\Leftrightarrow \sqrt{x}+1=3$
$\Leftrightarrow \sqrt{x}=2$
$\Leftrightarrow x=4$ (TM)
Vậy $x=4$