1,

`A=((\sqrt{x}-1)/(x-4)-(\sqrt{x}+1)/(x+4\sqrt{x}+4)):(x\sqrt{x})/((4-x)^2)(x>0; x\ne4)`

`=[(\sqrt{x}-1)/((\sqrt{x})^2-2^2)-(\sqrt{x}+1)/(x+2.\sqrt{x}.2+2^2)]:(x\sqrt{x})/((4-x)^2)`

`=[(\sqrt{x}-1)/((\sqrt{x}-2)(\sqrt{x}+2))-(\sqrt{x}+1)/((\sqrt{x}+1)^2)]:(x\sqrt{x})/((4-x)^2)`

`=[((\sqrt{x}-1)(\sqrt{x}+2))/((\sqrt{x}-2)(\sqrt{x}+2)^2)-((\sqrt{x}+1)(\sqrt{x}-2))/((\sqrt{x}-2)(\sqrt{x}+2)^2)]:(x\sqrt{x})/((4-x)^2)`

`=[((\sqrt{x}-1)(\sqrt{x}+2)-(\sqrt{x}+1)(\sqrt{x}-2))/((\sqrt{x}-2)(\sqrt{x}+2)^2)]:(x\sqrt{x})/((4-x)^2)`

`=[(x+2\sqrt{x}-\sqrt{x}-2-(x-2\sqrt{x}+\sqrt{x}-2))/((\sqrt{x}-2)(\sqrt{x}+2)^2)]:(x\sqrt{x})/((4-x)^2)`

`=[(x+2\sqrt{x}-\sqrt{x}-2-x+2\sqrt{x}-\sqrt{x}+2)/((\sqrt{x}-2)(\sqrt{x}+2)^2)]:(x\sqrt{x})/((4-x)^2)`

`=[(x+\sqrt{x}-2-x+\sqrt{x}+2)/((\sqrt{x}-2)(\sqrt{x}+2)^2)]:(x\sqrt{x})/((4-x)^2)`

`=[(2\sqrt{x})/((\sqrt{x}-2)(\sqrt{x}+2)^2)]:(x\sqrt{x})/((4-x)^2)`

`=[(2\sqrt{x})/((\sqrt{x}-2)(\sqrt{x}+2)^2)] . ((4-x)^2)/(x\sqrt{x})`

`=[(2\sqrt{x})/((\sqrt{x}-2)(\sqrt{x}+2)^2)] . ([2^2-(\sqrt{x})^2]^2)/(x\sqrt{x})`

`=[(2\sqrt{x})/((\sqrt{x}-2)(\sqrt{x}+2)^2)] . ([(2-\sqrt{x})(2+\sqrt{x})]^2)/(x\sqrt{x})`

`=(2\sqrt{x}.(2-\sqrt{x})^2 . (2+\sqrt{x})^2)/((\sqrt{x}-2).(\sqrt{x}+2)^2.x\sqrt{x})`

`=(2(\sqrt{x}-2))/x`

Vậy, `A=(2(\sqrt{x}-2))/x.`

2,

`x=4+2\sqrt{3}`

`->x=3+2\sqrt{3}+1`

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

`->\sqrt{x}=\sqrt{3}+1`

Với `\sqrt{x}=\sqrt{3}+1` và `x=4+2\sqrt{3}`

`A=(2(\sqrt{3}+1-2))/(4+2\sqrt{3})`

`->A=(2(\sqrt{3}-1))/(4+2\sqrt{3})`

`->A=(2\sqrt{3}-2)/(4+2\sqrt{3})`

`->A=((2\sqrt{3}-2)(4-2\sqrt{3}))/((4+2\sqrt{3})(4-2\sqrt{3}))`

`->A=(8\sqrt{3}-12-8+4\sqrt{3})/(4^2-(2\sqrt{3})^2)`

`->A=(12\sqrt{3}-20)/(16-12)`

`->A=(12\sqrt{3}-20)/4`

`->A=(12\sqrt{3})/4-20/4`

`->A=3\sqrt{3}-5`

Vậy, `A=3\sqrt{3}-5` với `\sqrt{x}=\sqrt{3}+1` và `x=4+2\sqrt{3}.`

3,

Xét `A-1/4`

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

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

`=(8(\sqrt{x}-2))/(4x)-x/(4x)`

`=(8\sqrt{x}-16-x)/(4x)`

`=-(-8\sqrt{x}+16+x)/(4x)`

`=-([(\sqrt{x})^2-2.\sqrt{x}.4+4^2])/(4x)`

`=-((\sqrt{x}-4)^2)/x`

Vì:

`(\sqrt{x}-4)^2>=0 AA x`

`-> -(\sqrt{x}-4)^2<=0 AA x`

`->-(\sqrt{x}-4)^2<=0 AA`

Vì thế:

`\sqrt{x}-4=0`

`<=>\sqrt{x}=4`

`<=>x=16`(Tm)

Vậy, `x=16` thì `A>=1/4.`