Đáp án: `P=\sqrt{x}-2` với $x≥0;x\neq1$
Giải thích các bước giải:
`P=\frac{\sqrt{x}}{\sqrt{x}+2}+\frac{-x+x\sqrt{x}+6}{x+\sqrt{x}-2}-\frac{\sqrt{x}+1}{\sqrt{x}-1}`
`=\frac{\sqrt{x}(\sqrt{x}-1)+(x\sqrt{x}-x+6)-(\sqrt{x}+1)(\sqrt{x}+2)}{(\sqrt{x}+2)(\sqrt{x}-1)}`
`=\frac{(x-\sqrt{x})+(x\sqrt{x}-x+6)-(x+3\sqrt{x}+2)}{(\sqrt{x}+2)(\sqrt{x}-1)}`
`=\frac{x-\sqrt{x}+x\sqrt{x}-x+6-x-3\sqrt{x}-2}{(\sqrt{x}+2)(\sqrt{x}-1)}`
`=\frac{x\sqrt{x}-x-4\sqrt{x}+4}{(\sqrt{x}+2)(\sqrt{x}-1)}`
`=\frac{x(\sqrt{x}-1)-4(\sqrt{x}-1)}{(\sqrt{x}+2)(\sqrt{x}-1)}`
`=\frac{(\sqrt{x}-1)(x-4)}{(\sqrt{x}+2)(\sqrt{x}-1)}`
`=\frac{(\sqrt{x}-1)(\sqrt{x}-2)(\sqrt{x}+2)}{(\sqrt{x}+2)(\sqrt{x}-1)}`
`=\sqrt{x}-2`
