`a)`
Ta có:`x+\sqrt{x}-2=x-\sqrt{x}+2\sqrt{x}-2`
`=\sqrt{x}(\sqrt{x}-1)+2(\sqrt{x}-1)`
`(\sqrt{x}+2)(\sqrt{x}-1)`
`P={3x+\sqrt{9x}-3}/{x+\sqrt{x}-2}-{\sqrt{x}+1}/{\sqrt{x}+2}+{\sqrt{x}-2}/{1-\sqrt{x}}` $(x\ge 0; x\ne 1)$
`P={3x+3\sqrt{x}-3}/{(\sqrt{x}+2)(\sqrt{x}-1)}-{(\sqrt{x}+1)(\sqrt{x}-1)}/{(\sqrt{x}+2)(\sqrt{x}-1)}-{(\sqrt{x}-2)(\sqrt{x}+2)}/{(\sqrt{x}+2)(\sqrt{x}-1)}`
`P={3x+3\sqrt{x}-3-(x-1)-(x-4)}/{(\sqrt{x}+2)(\sqrt{x}-1)}`
`P={x+3\sqrt{x}+2}/{(\sqrt{x}+2)(\sqrt{x}-1)}`
`P={x+2\sqrt{x}+\sqrt{x}+2}/{(\sqrt{x}+2)(\sqrt{x}-2)}`
`P={(\sqrt{x}+1)(\sqrt{x}+2)}/{(\sqrt{x}+2)(\sqrt{x}-1)}`
`P={\sqrt{x}+1}/{\sqrt{x}-1}`
`b)` `P={\sqrt{x}+1}/{\sqrt{x}-1}={(\sqrt{x}+1)(\sqrt{x}+1)}/{(\sqrt{x}-1)(\sqrt{x}+1)}`
`P={(\sqrt{x}+1)^2}/{x-1}`
`\sqrt{P}` có nghĩa khi $x-1>0\Leftrightarrow x>1$
+) `\sqrt{P}=\sqrt{{(\sqrt{x}+1)^2}/{x-1}}={\sqrt{x}+1}/ \sqrt{x-1}` $(x>1)$
+) `P-\sqrt{P}={\sqrt{x}+1}/{\sqrt{x}-1}-{\sqrt{x}+1}/ \sqrt{x-1}`
`=(\sqrt{x}+1)(1/{\sqrt{x}-1}-1/{\sqrt{x-1}})`
+) `x>1=>\sqrt{x}>1=>2\sqrt{x}>2=>-2\sqrt{x}<-2`
`=>x-2\sqrt{x}+1<x-2+1`
`=>(\sqrt{x}-1)^2<x-1=(\sqrt{x-1})^2`
`=>\sqrt{x}-1<\sqrt{x-1}`
`=>1/{\sqrt{x}-1}>1/{\sqrt{x-1}}`
`=>1/{\sqrt{x}-1}-1/{\sqrt{x-1}} >0`
`=>P-\sqrt{P}=(\sqrt{x}+1)(1/{\sqrt{x}-1}-1/{\sqrt{x-1}})>0`
`=>P>\sqrt{P}` với $x>1$
`c)`
`1/P={\sqrt{x}-1}/{\sqrt{x}+1}={\sqrt{x}+1-2}/{\sqrt{x}+1}=1-2/{\sqrt{x}+1`
Để `1/P \in Z=>(\sqrt{x}+1)\in Ư(2)={-2;-1;1;2}`
Mà `\sqrt{x}+1\ge 1 \forall x\ge 0`
`=>\sqrt{x}+1\in {1;2}`
`=>\sqrt{x}\in {0;1}`
`=>x\in {0;1}`
Kết hợp điều kiện `x\ne 1=>x=0`
Vậy $x=0$ thì `1/P \in Z`
