Bạn tham khảo nhé.
`A=\frac{\sqrt{x}}{\sqrt{x}+1}(x>=0)`
Khi `A=\sqrt{x}-2`
`=>\frac{\sqrt{x}}{\sqrt{x}+1}=\sqrt{x}-2`
`<=>\sqrt{x}=(\sqrt{x}-2)(\sqrt{x}+1)`
`<=>\sqrt{x}=x+\sqrt{x}-2\sqrt{x}-2`
`<=>\sqrt{x}=x-\sqrt{x}-2`
`<=>x-\sqrt{x}-\sqrt{x}-2=0`
`<=>x-2\sqrt{x}+1-3=0`
`<=>(\sqrt{x}-1)^2=3`
`<=>(\sqrt{x}-1)^2=(+-\sqrt{3})^2`
`<=>\sqrt{x}-1=+-\sqrt{3}`
`<=>\sqrt{x}=+-\sqrt{3}+1`
`=>[(x=(\sqrt{3}+1)^2),(x=(-\sqrt{3}+1)^2):}`
`<=>[(x=3+2\sqrt{3}+1),(x=1-2\sqrt{3}+3):}`
`<=>[(x=4+2\sqrt{3}),(x=4-2\sqrt{3}):}`
Vậy `x\in{4+-2\sqrt{3}}` khi `A=\sqrt{x}-2`
