ĐK : tự tìm
$A = \dfrac{x\sqrt{x}-3}{x-2\sqrt{x}-3} - \dfrac{2(\sqrt{x}-3)}{\sqrt{x}+1} + \dfrac{\sqrt{x}+3}{3-\sqrt{x}}$
$= \dfrac{x\sqrt{x} - 3}{(\sqrt{x}-3)(\sqrt{x}+1)} - \dfrac{2(\sqrt{x}-3)^2}{(\sqrt{x}-3)(\sqrt{x}+1)} - \dfrac{(\sqrt{x}+3)(\sqrt{x}+1)}{(\sqrt{x}-3)(\sqrt{x}+1)}$
$= \dfrac{x\sqrt{x}-3 - 2x + 12\sqrt{x} - 18 - x - 4\sqrt{x} - 3}{(\sqrt{x}-3)(\sqrt{x}+1)}$
$= \dfrac{x\sqrt{x} - 3x + 8\sqrt{x} - 24}{(\sqrt{x}-3)(\sqrt{x}+1)}$
$= \dfrac{x(\sqrt{x} - 3) + 8(\sqrt{x} -3)}{(\sqrt{x}-3)(\sqrt{x}+1)}$
$= \dfrac{(\sqrt{x}-3)(x+8)}{(\sqrt{x}-3)(\sqrt{x}+1)}$
$= \dfrac{x+8}{\sqrt{x}+1}$
$= \dfrac{x-1+9}{\sqrt{x}+1}$
$= \dfrac{(\sqrt{x}+1)(\sqrt{x}-1)+9}{\sqrt{x}+1}$
$= \sqrt{x}-1 + \dfrac9{\sqrt{x}+1}$
$= \sqrt{x}+1 + \dfrac9{\sqrt{x}+1} - 2$
$\overset{AM-GM}{\geqslant} 2\sqrt{(\sqrt{x}+1)\dfrac9{\sqrt{x}+1}} - 2$
$= 2.\sqrt{9} - 2 = 4$
Dấu '=' xảy ra $\iff \sqrt{x}+1 = \dfrac9{\sqrt{x}+1} \iff x = 4$
Vậy $A_\textrm{min} = 4 \iff x = 4$