a, $A=(\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}-2}):\dfrac{\sqrt{x}}{x-2\sqrt{x}}$ $(x>0;x\neq4)$
$=\dfrac{\sqrt{x}-2-\sqrt{x}-2}{(\sqrt{x}+2)(\sqrt{x}-2)}.\dfrac{\sqrt{x}(\sqrt{x}-2)}{\sqrt{x}}$
$=\dfrac{-4}{\sqrt{x}+2}$
b, $A=\dfrac{-2}{3}$
$⇔\dfrac{-4}{\sqrt{x}+2}=\dfrac{-2}{3}$
$⇔\sqrt{x}+2=6$
$⇔\sqrt{x}=4$
$⇔x=16(tm)$
c, Để $A_{Min}$, mà $-4<0$
$⇒(\sqrt{x}+2)_{Min}$ $(x\neq0)$
$⇒\sqrt{x}=1$
$⇒x=1$
Khi đó $A_{Min}=\dfrac{-4}{1+2}=\dfrac{-4}{3}$