ĐKXĐ: $x> 0,x\ne 4,x\ne 9$
$P\,=\left(\dfrac{\sqrt x-3}{2-\sqrt x}+\dfrac{\sqrt x+2}{3+\sqrt x}-\dfrac{9-x}{x+\sqrt x-6}\right):\left(1-\dfrac{3\sqrt x-9}{x-9}\right)\\\quad =\left(\dfrac{-(\sqrt x-3)(\sqrt x+3)}{(\sqrt x-2)(\sqrt x+3)}+\dfrac{(\sqrt x+2)(\sqrt x-2)}{(\sqrt x+3)(\sqrt x-2)}+\dfrac{x-9}{x+3\sqrt x-2\sqrt x-6}\right):\left(1-\dfrac{3(\sqrt x-3)}{(\sqrt x-3)(\sqrt x+3)}\right)\\\quad =\left(\dfrac{-(x-9)+(x-4)}{(\sqrt x+3)(\sqrt x-2)}+\dfrac{x-9}{(x+3\sqrt x)-(2\sqrt x+6)}\right):\left(\dfrac{\sqrt x+3}{\sqrt x+3}-\dfrac{3}{\sqrt x+3}\right)\\\quad =\left(\dfrac{-x+9+x-4}{(\sqrt x+3)(\sqrt x-2)}+\dfrac{x-9}{\sqrt x(\sqrt x+3)-2(\sqrt x+3)}\right):\dfrac{\sqrt x+3-3}{\sqrt x+3}\\\quad =\left(\dfrac{5}{(\sqrt x+3)(\sqrt x-2)}+\dfrac{x-9}{(\sqrt x+3)(\sqrt x-2)}\right).\dfrac{\sqrt x+3}{\sqrt x}\\\quad =\dfrac{5+x-9}{\sqrt x-2}.\dfrac{1}{\sqrt x}\\\quad =\dfrac{x-4}{\sqrt x-2}.\dfrac{1}{\sqrt x}\\\quad =\dfrac{(\sqrt x-2)(\sqrt x+2)}{\sqrt x-2}.\dfrac{1}{\sqrt x}\\\quad =\dfrac{\sqrt x+2}{\sqrt x}$
Vậy $P=\dfrac{\sqrt x+2}{\sqrt x}$ với $x>0,\,x\ne 4,\,x\ne 9$
