`a)A=\frac{x\sqrt{x}+1}{x+2\sqrt{x}+1}(x>=0)`
`=\frac{(\sqrt{x}+1)(x-\sqrt{x}+1)}{(\sqrt{x}+1)^2}`
`=\frac{x-\sqrt{x}+1}{\sqrt{x}+1}`
`x=9+4\sqrt{5}`
`=>\sqrt{x}=\sqrt{9+4\sqrt{5}}=\sqrt{(\sqrt{5}+2)^2}=\sqrt{5}+2`
`->`Thế `x` và `\sqrt{x}` vào biểu thức
`->A=\frac{9+4\sqrt{5}-\sqrt{5}-2+1}{\sqrt{5}+2+1}`
`=\frac{8+3\sqrt{5}}{\sqrt{5}+3}`
`=\frac{(8+3\sqrt{5})(\sqrt{5}-3)}{5-9}`
`=\frac{8\sqrt{5}-21+15-9\sqrt{5}}{-4}`
`=\frac{-\sqrt{5}-6}{-4}`
`=\frac{\sqrt{5}+6}{4}`
Vậy `A=\frac{\sqrt{5}+6}{4}` khi `x=9+4\sqrt{5}`
`b)M=A.B`
`=\frac{x-\sqrt{x}+1}{\sqrt{x}+1}.(\frac{2x+6\sqrt{x}+7}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1})`
`=\frac{x-\sqrt{x}+1}{\sqrt{x}+1}.(\frac{2x+6\sqrt{x}+7-x+\sqrt{x}-1}{(\sqrt{x}+1)(x-\sqrt{x}+1)})`
`=\frac{x-\sqrt{x}+1}{\sqrt{x}+1}.\frac{x+7\sqrt{x}+6}{(\sqrt{x}+1)(x-\sqrt{x}+1)}`
`=\frac{(x-\sqrt{x}+1)(\sqrt{x}+1)(\sqrt{x}+6)}{(\sqrt{x}+1)(\sqrt{x}+1)(x-\sqrt{x}+1)}`
`=\frac{\sqrt{x}+6}{\sqrt{x}+1}`
`M>2`
`<=>\frac{\sqrt{x}+6}{\sqrt{x}+1}>2`
`<=>\frac{\sqrt{x}+6-\sqrt{x}-1}{\sqrt{x}+1}>0`
`<=>\frac{5}{\sqrt{x}+1}>0`
`<=>\sqrt{x}+1>0`(luôn đúng)
Vậy với mọi `x>0` thì `M>2`
`c)M=\frac{\sqrt{x}+6}{\sqrt{x}+1}`
Để `M\in ZZ` thì `\sqrt{x}+6 \vdots \sqrt{x}+1`
`<=>5 \vdots \sqrt{x}+1`
`<=>\sqrt{x}+1 \in Ư(5)`
`->\sqrt{x}+1 \in {±1;±5}`
Vì `\sqrt{x}>=0<=>\sqrt{x}+1>=1`
`<=>\sqrt{x}+1=1<=>x=0`
`<=>\sqrt{x}+1=5<=>x=16`
Vậy `x\in{0;16}` thì `M\in ZZ`
