Đáp án:
`a)`
Với `x>=0;x\ne9` ta có:
`A=(\sqrt{x})/(\sqrt{x}+3)+(2\sqrt{x})/(\sqrt{x}-3)-(3x+9)/(x-9)`
`=(\sqrt{x})/(\sqrt{x}+3)+(2\sqrt{x})/(\sqrt{x}-3)-(3x+9)/((\sqrt{x}-3).(\sqrt{x}+3))`
`=(\sqrt{x}.(\sqrt{x}-3))/((\sqrt{x}+3).(\sqrt{x}-3))+(2\sqrt{x}.(\sqrt{x}+3))/((\sqrt{x}-3).(\sqrt{x}+3))-(3x+9)/((\sqrt{x}-3).(\sqrt{x}+3))`
`=(x-3\sqrt{x}+2x+6\sqrt{x}-3x-9)/((\sqrt{x}-3).(\sqrt{x}+3))`
`=(3\sqrt{x}-9)/((\sqrt{x}-3).(\sqrt{x}+3))`
`=(3.(\sqrt{x}-3))/((\sqrt{x}-3).(\sqrt{x}+3))`
`=(3)/(\sqrt{x}+3)`
Vậy `A=(3)/(\sqrt{x}+3)` với `x>=0;x\ne9`
`b)`
`A=1/3`
`<=>(3)/(\sqrt{x}+3)=1/3(x>=0;x\ne9)`
`<=>\sqrt{x}+3=9`
`<=>\sqrt{x}=6`
`<=>x=6^2`
`<=>x=36(tmđk)`
Vậy `x=36` thì `A=1/3`
`c)`
Với `x>=0`
`<=>\sqrt{x}>=0`
`<=>\sqrt{x}+3>=3`
`<=>(1)/(\sqrt{x}+3)<=(1)/(3)`
`<=>(3)/(\sqrt{x}+3)<=3/3=1`
`->A<=1`
Dấu `'='` xảy ra `<=>x=0`
Vậy `A_{max}=1` khi `x=0`
