Cho A = $\frac{√x}{√x + 1}$ với x $\geq$ 0. Khi A = √x - 2 thì x bằng bao nhiêu?

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`