Giải thích các bước giải:
$AB=\dfrac{x-4}{\sqrt{x}-1}.\dfrac{1}{\sqrt{x}-2}$
$\to AB=\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)}{\sqrt{x}-1}.\dfrac{1}{\sqrt{x}-2}$
$\to AB=\dfrac{\sqrt{x}+2}{\sqrt{x}-1}$
$\to \dfrac{18}{AB}=\dfrac{18(\sqrt{x}-1)}{\sqrt{x}+2}$
$\to \dfrac{18}{AB}=\dfrac{18(\sqrt{x}+2)-54}{\sqrt{x}+2}$
$\to \dfrac{18}{AB}=18-\dfrac{54}{\sqrt{x}+2}$
$\to \dfrac{18}{AB}\ge 18-\dfrac{54}{0+2}=-9$
Dấu = xảy ra khi $x=0$