`a)x-5\sqrt{x}+6=(\sqrt{x}-3)(\sqrt{x}-2)`
Để `A` có nghĩa
`<=>{(x>=0),(\sqrt{x}-3\ne0),(\sqrt{x}-2\ne0):}`
`<=>{(x>=0),(\sqrt{x}\ne3),(\sqrt{x}\ne2):}`
`<=>{(x>=0),(x\ne9),(x\ne4):}`
Vậy `x>=0; x\ne9; x\ne4` thì `A` có nghĩa
`A=(\sqrt{x}+2)/(\sqrt{x}-3)-(\sqrt{x}+1)/(\sqrt{x}-2)+(3-3\sqrt{x})/x-(5\sqrt{x}+6)`
`A=(\sqrt{x}+2)/(\sqrt{x}-3)-(\sqrt{x}+1)/(\sqrt{x}-2)+(3-3\sqrt{x})/x-((\sqrt{x}-3)(\sqrt{x}-2))`
`A=((\sqrt{x}+2)(\sqrt{x}-2)-(\sqrt{x}+1)(\sqrt{x}-3)+(3-3\sqrt{x}))/((\sqrt{x}-3)(\sqrt{x}-2))`
`A=(x-4-x+3\sqrt{x}-\sqrt{x}+3+3-3\sqrt{x})/((\sqrt{x}-3)(\sqrt{x}-2))`
`A=(-\sqrt{x}+2)/((\sqrt{x}-3)(\sqrt{x}-2))`
`A=(-(\sqrt{x}-2))/((\sqrt{x}-3)(\sqrt{x}-2))`
`A=-1/(\sqrt{x}-3)`
Vậy với `x>=0; x\ne9; x\ne4` thì `A=-1/(\sqrt{x}-3)`
`b)`Để `A>1`
`<=> -1/(\sqrt{x}-3)>1`
`<=> -1/(\sqrt{x}-3)-1>0`
`<=>(-1-(\sqrt{x}-3))/(\sqrt{x}-3)>0`
`<=>(-1-\sqrt{x}+3)/(\sqrt{x}-3)>0`
`<=>(2-\sqrt{x})/(\sqrt{x}-3)>0`
`<=>[({(2-\sqrt{x}>0),(\sqrt{x}-3):}),({(2-\sqrt{x}<0),(\sqrt{x}-3<0):}):}`
`<=>[({(-\sqrt{x}> -2),(\sqrt{x}>3):}),({(-\sqrt{x}<-2),(\sqrt{x}<3):}):}`
`<=>[({(\sqrt{x}<2),(x>9):}),({(\sqrt{x}>2),(x<9):}):}`
`<=>[({(x<4),(x>9):}(L)),({(x>4),(x<9):}(TM)):}`
`<=>4<x<9`
Vậy `4<x<9` thì `A>1`
`c)`Để `A` đạt `GTN N`
`<=> -1/(\sqrt{x}-3)` đạt `GTN N`
`<=>1/(\sqrt{x}-3)` đạt `GTLN`
`<=>\sqrt{x}-3` đạt `GTN N`
`<=>\sqrt{x}-3>=-3`
Dấu `=` xảy ra `<=>x=0(TM)`
`=>A>=(-1)/(-3)=1/3`
Vậy `min A=1/3` khi `x=0`
