`a)A=(\sqrt{x}+1)/(x+4\sqrt{x}+4):(x/(x+2\sqrt{x})+x/(\sqrt{x}+2))(x>0)`

`=(\sqrt{x}+1)/(x+4\sqrt{x}+4):(x/(\sqrt{x}(\sqrt{x}+2))+x/(\sqrt{x}+2))`

`=(\sqrt{x}+1)/(x+4\sqrt{x}+4):(\sqrt{x}/(\sqrt{x}+2)+x/(\sqrt{x}+2))`

`=(\sqrt{x}+1)/(\sqrt{x}+2)^2:(\sqrt{x}+x)/(\sqrt{x}+2)`

`=(\sqrt{x}+1)/(\sqrt{x}+2)^2*(\sqrt{x}+2)/(x+\sqrt{x})`

`=(\sqrt{x}+1)/(\sqrt{x}+2)^2*(\sqrt{x}+2)/(\sqrt{x}(\sqrt{x}+1))`

`=1/(\sqrt{x}(\sqrt{x}+2})`

`=1/(x+2\sqrt{x})`

`b)A>=1/(3\sqrt{x})`

Suy ra: `1/(x+2\sqrt{x})>=1/(3\sqrt{x})`

Vì: `x>0->\sqrt{x}>0`

Do đó ta nhân `\sqrt{x}` cho 2 vế ta được:

`\sqrt{x}*1/(x+2\sqrt{x})>=1/(3\sqrt{x})*\sqrt{x}`

`1/(\sqrt{x}+2)>=1/3`

`1>=1/3(\sqrt{x}+2)` (vì `\sqrt{x}+2>0)`

`\sqrt{x}+2<=3`

`\sqrt{x}<=3-2=1`

`x<=1^2=1`

Kết hợp với đk thì: `0<x<=1`

Vậy: `0<x<=1`