`P=\frac{1-\sqrt{x}}{x-\sqrt{x}+1}.\frac{x\sqrt{x}+1}{x+2\sqrt{x}+1}` ĐK: x $\geq$ 0
`P=\frac{1-\sqrt{x}}{x-\sqrt{x}+1}.\frac{(\sqrt{x}+1)(x-\sqrt{x}+1)}{(\sqrt{x}+1)^2}`
`P=\frac{1-\sqrt{x}}{\sqrt{x}+1}`
Vậy `P=\frac{1-\sqrt{x}}{\sqrt{x}+1}` với x $\geq$ 0