`\qquad A=(2\sqrt{x}+1)/(\sqrt{x}-3)+(2\sqrt{x}-9)/(x-5\sqrt{x}+6)-(\sqrt{x}+3)/(\sqrt{x}-2)`
`a) \text{ĐKXĐ:}` `{(x>=0),(\sqrt{x}-3\ne0),(\sqrt{x}-2\ne0),(x-5\sqrt{x}+6\ne0):}<=>{(x>=0),(\sqrt{x}\ne3),(\sqrt{x}\ne2):}<=>{(x>=0),(x\ne9),(x\ne4):}`
Với `x>=0,x\ne9,x\ne4` thì
`A=((2\sqrt{x}+1)(\sqrt{x}-2)+2\sqrt{x}-9-(\sqrt{x}-3)(\sqrt{x}+3))/((\sqrt{x}-3)(\sqrt{x}-2))`
`A=(2x-4\sqrt{x}+\sqrt{x}-2+2\sqrt{x}-9-x+9)/((\sqrt{x}-3)(\sqrt{x}-2))`
`A=(x-\sqrt{x}-2)/((\sqrt{x}-3)(\sqrt{x}-2))`
`A=(x-2\sqrt{x}+\sqrt{x}-2)/((\sqrt{x}-3)(\sqrt{x}-2))`
`A=((\sqrt{x}-2)(\sqrt{x}+1))/((\sqrt{x}-3)(\sqrt{x}-2))`
`A=(\sqrt{x}+1)/(\sqrt{x}-3)`
Vậy `A=(\sqrt{x}+1)/(\sqrt{x}-3)` với `x>=0;x\ne9;x\ne4`
`b) x=6+2\sqrt{5}->\sqrt{x}=\sqrt{6+2\sqrt{5}}=\sqrt{(\sqrt{5}+1)^2}=\sqrt{5}+1`
Thay `\sqrt{x}=\sqrt{5}+1` vào A ta có:
`A=(\sqrt{5}+1+1)/(\sqrt{5}+1-3)=(\sqrt{5}+2)/(\sqrt{5}-2)=(\sqrt{5}+2)^2/(5-4)=9+4\sqrt{5}`
`c) A=(\sqrt{x}+1)/(\sqrt{x}-3)=3`
`<=> \sqrt{x}+1=3(\sqrt{x}-3)`
`<=> \sqrt{x}+1=3\sqrt{x}-9`
`<=> 2\sqrt{x}=10 `
`<=>\sqrt{x}=5`
`<=> x=25(\text{tm})`
Vậy x=25 thỏa mãn `A=3`
