Ta có:
`\qquad x-5\sqrt{x}+6=x-2\sqrt{x}-3\sqrt{x}+2.3`
`=\sqrt{x}.(\sqrt{x}-2)-3.(\sqrt{x}-2)`
`=(\sqrt{x}-2)(\sqrt{x}-3)`
Suy ra:
`\qquad {2\sqrt{x}-9}/{x-5\sqrt{x}+6} - {\sqrt{x}+3}/{\sqrt{x}-2}+{2\sqrt{x}+1}/{\sqrt{x}-3}` `(x\ne 4;x \ne 9)`
`={2\sqrt{x}-9-(\sqrt{x}+3)(\sqrt{x}-3)+(2\sqrt{x}+1)(\sqrt{x}-2)}/{(\sqrt{x}-2)(\sqrt{x}-3)}`
`={2\sqrt{x}-9-(x-9)+2x-4\sqrt{x}+\sqrt{x}-2}/{(\sqrt{x}-2)(\sqrt{x}-3)}`
`={x-\sqrt{x}-2}/{(\sqrt{x}-2)(\sqrt{x}-3)}`
`={x-2\sqrt{x}+\sqrt{x}-2}/{(\sqrt{x}-2)(\sqrt{x}-3)}`
`={\sqrt{x}(\sqrt{x}-2)+\sqrt{x}-2}/{(\sqrt{x}-2)(\sqrt{x}-3)}`
`={(\sqrt{x}-2)(\sqrt{x}+1)}/{(\sqrt{x}-2)(\sqrt{x}-3)}`
`={\sqrt{x}+1}/{\sqrt{x}-3}`
Vậy:
`{2\sqrt{x}-9}/{x-5\sqrt{x}+6} - {\sqrt{x}+3}/{\sqrt{x}-2}+{2\sqrt{x}+1}/{\sqrt{x}-3}={\sqrt{x}+1}/{\sqrt{x}-3}\quad (đpcm)`
