$a)P=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\\ =\dfrac{3x+\sqrt{9x}-3}{(\sqrt{x}+2)(\sqrt{x}-1)}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\\ =\dfrac{3x+\sqrt{9x}-3}{(\sqrt{x}+2)(\sqrt{x}-1)}-\dfrac{(\sqrt{x}+1)(\sqrt{x}-1)}{(\sqrt{x}+2)(\sqrt{x}-1)}-\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)}{(\sqrt{x}+2)(\sqrt{x}-1)}\\ =\dfrac{3x+\sqrt{9x}-3-(\sqrt{x}+1)(\sqrt{x}-1)-(\sqrt{x}-2)(\sqrt{x}+2)}{(\sqrt{x}+2)(\sqrt{x}-1)}\\ =\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{(\sqrt{x}+2)(\sqrt{x}-1)}\\ =\dfrac{x+3\sqrt{x}+2}{(\sqrt{x}+2)(\sqrt{x}-1)}\\ =\dfrac{(\sqrt{x}+2)(\sqrt{x}+1)}{(\sqrt{x}+2)(\sqrt{x}-1)}\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\\ b)\sqrt{P}=\sqrt{\dfrac{\sqrt{x}+1}{\sqrt{x}-1}}(ĐK:x>1)\\ \Rightarrow \sqrt{x}+1 > \sqrt{x}-1;\sqrt{x}+1;\sqrt{x}+1 >0\\ \Rightarrow P=\dfrac{\sqrt{x}+1}{\sqrt{x}-1} >1\\ \Rightarrow P>\sqrt{P}\\ c) \dfrac{1}{P}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1-2}{\sqrt{x}+1}=1-\dfrac{2}{\sqrt{x}+1}$
Để $P$ nguyên thì $2 \vdots (\sqrt{x}+1)$
$\Rightarrow x \in \{0;1\}$
