A=$\frac{2x+1}{x√x-1}$ -$\frac{1}{ √x-1}$ B=$\frac{ √x+3}{x+√x+1}$ Đặt P=A:B.Rút gọn P

Đáp án+Giải thích các bước giải:

`A=\frac{2x+1}{x\sqrt{x}-1}-\frac{1}{\sqrt{x}-1}(x≥0;x\ne1)`

`A=\frac{2x+1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}-\frac{x+\sqrt{x}+1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}`

`A=\frac{2x+1-x-\sqrt{x}-1}{(\sqrt{x}-1)(x+\sqrt{x}+1)}`

`A=\frac{x-\sqrt{x}}{(\sqrt{x}-1)(x+\sqrt{x}+1)}`

`A=\frac{\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}-1)(x+\sqrt{x}+1)}`

`A=\frac{\sqrt{x}}{x+\sqrt{x}+1}`

`P=A:B`

`⇔P=\frac{\sqrt{x}}{x+\sqrt{x}+1}:\frac{\sqrt{x}+3}{x+\sqrt{x}+1}`

`⇔P=\frac{\sqrt{x}}{x+\sqrt{x}+1}.\frac{x+\sqrt{x}+1}{\sqrt{x}+3}`

`⇔P=\frac{\sqrt{x}}{\sqrt{x}+3}`

Vậy với `x≥0;\xne1` thì `P=\frac{\sqrt{x}}{\sqrt{x}+3}`