`a)` Ta có:
`\qquad 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)^2}/{(\sqrt{x}+2)(\sqrt{x}-1)}`
`P={3x+3\sqrt{x}-3-(x-1)-(x+4\sqrt{x}+4)}/{(\sqrt{x}+2)(\sqrt{x}-1)}`
`P={x-\sqrt{x}-6}/{(\sqrt{x}+2)(\sqrt{x}-1)}`
`P={x-3\sqrt{x}+2\sqrt{x}-6}/{(\sqrt{x}+2)(\sqrt{x}-1)}`
`P={\sqrt{x}(\sqrt{x}-3)+2(\sqrt{x}-3)}/{(\sqrt{x}+2)(\sqrt{x}-1)}`
`P={(\sqrt{x}-3)(\sqrt{x}+2)}/{(\sqrt{x}+2)(\sqrt{x}-1)}`
`P={\sqrt{x}-3}/{\sqrt{x}-1}`
`b)` `\qquad P=2`
`<=>{\sqrt{x}-3}/{\sqrt{x}-1}=2`
`<=>\sqrt{x}-3=2(\sqrt{x}-1)`
`<=>2\sqrt{x}-2-\sqrt{x}+3=0`
`<=>\sqrt{x}=-1` (vô nghiệm)
Vậy không có giá trị $x$ để $P=2$
