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