Giải thích các bước giải:
ĐKXĐ: $x\ge 0, x\ne 9, 4$
Ta có:
$P=(\dfrac{2\sqrt{x}}{\sqrt{x}+3}-\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}):(\dfrac{2}{\sqrt{x}-2}-\dfrac{1}{\sqrt{x}-3})$
$\to P=(\dfrac{2\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}+3)(\sqrt{x}-3)}-\dfrac{\sqrt{x}(\sqrt{x}+3)}{(\sqrt{x}+3)(\sqrt{x}-3)}-\dfrac{3x+3}{(\sqrt{x}+3)(\sqrt{x}-3)}):\dfrac{2(\sqrt{x}-3)-1(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}-3)}$
$\to P=\dfrac{2\sqrt{x}(\sqrt{x}-3)-\sqrt{x}(\sqrt{x}+3)-(3x+3)}{(\sqrt{x}+3)(\sqrt{x}-3)}:\dfrac{\sqrt{x}-4}{(\sqrt{x}-2)(\sqrt{x}-3)}$
$\to P=\dfrac{-2x-9\sqrt{x}-3}{(\sqrt{x}+3)(\sqrt{x}-3)}\cdot\dfrac{(\sqrt{x}-2)(\sqrt{x}-3)}{\sqrt{x}-4}$
$\to P=\dfrac{(-2x-9\sqrt{x}-3)(\sqrt{x}-2)}{(\sqrt{x}+3)(\sqrt{x}-4)}$
