\(P=\dfrac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}(\)ĐKXĐ:\(x>1)\)
\(=\dfrac{3x+3\sqrt{x}-3}{x-\sqrt{x}+2\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\dfrac{3x+3\sqrt{x}-3}{\sqrt{x}(\sqrt{x}-1)+2(\sqrt{x}-1)}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\dfrac{3x+3\sqrt{x}-3}{(\sqrt{x}+2)(\sqrt{x}-1)}-\dfrac{(\sqrt{x}+1)(\sqrt{x}-1)}{(\sqrt{x}+2)(\sqrt{x}-1)}-\dfrac{(\sqrt{x}-2)(\sqrt{x}+2)}{(\sqrt{x}+2)(\sqrt{x}-1)}\)
\(=\dfrac{3x+3\sqrt{x}-3-(\sqrt{x}+1)(\sqrt{x}-1)-(\sqrt{x}-2)(\sqrt{x}+2)}{(\sqrt{x}+2)(\sqrt{x}-1)}\)
\(=\dfrac{3x+3\sqrt{x}-3-x+1-x+4}{(\sqrt{x}+2)(\sqrt{x}-1)}\)
\(=\dfrac{x+3\sqrt{x}+2}{(\sqrt{x}+2)(\sqrt{x}-1)}\)
\(=\dfrac{x+\sqrt{x}+2\sqrt{x}+2}{(\sqrt{x}+2)(\sqrt{x}-1)}\)
\(=\dfrac{\sqrt{x}(\sqrt{x}+1)+2(\sqrt{x}+1)}{(\sqrt{x}+2)(\sqrt{x}-1)}\)
\(=\dfrac{(\sqrt{x}+2)(\sqrt{x}+1)}{(\sqrt{x}+2)(\sqrt{x}-1)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
Vậy \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\) với \(x>1\)
\(a)\)
 Để \(P=3\)
\(⇒\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=3\)
\(⇔\sqrt{x}+1=3(\sqrt{x}-1)\)
\(⇔\sqrt{x}+1=3\sqrt{x}-3\)
\(⇔-2\sqrt{x}=-4\)
\(⇔\sqrt{x}=2\)
\(⇔x=4\)(tmđk)
Vậy \(P=3⇒x=4\)
\(b)\)
\(P=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=1+\dfrac{2}{\sqrt{x}-1}\)
Để \(P∈\mathbb{Z}⇒\dfrac{2}{\sqrt{x}-1}∈\mathbb{Z}\)
\(⇒(\sqrt{x}-1)∈Ư(2)=\){\(±1;±2\)}
\(⇒(\sqrt{x}-1)∈\){\(-2;-1;1;2\)}
\(⇒\sqrt{x}∈\){\(-1;0;2;3\)}
Vì \(\sqrt{x}>1∀x>1⇒\sqrt{x}∈\){\(2;3\)}
\(⇒x∈\){\(4;9\)}
Kết hợp ĐK \(x>1⇒x∈\){\(4;9\)}
Vậy \(P∈\mathbb{Z}⇒x∈\){\(4;9\)}