Đáp án:
`P=x/{\sqrt{x}-1}` với `x>0;x\ne 1`
Giải thích các bước giải:
`P={x+\sqrt{x}}/{x-2\sqrt{x}+1}:({\sqrt{x}+1}/\sqrt{x}-1/{1-\sqrt{x}}+{2-x}/{x-\sqrt{x}})`
`\qquad (x>0; x\ne 1)`
`= {x+\sqrt{x}}/{(\sqrt{x}-1)^2}:({\sqrt{x}+1}/\sqrt{x}+1/{\sqrt{x}-1}+{2-x}/{\sqrt{x}(\sqrt{x}-1)})`
`={x+\sqrt{x}}/{(\sqrt{x}-1)^2}:{(\sqrt{x}+1)(\sqrt{x}-1)+\sqrt{x}+2-x}/{\sqrt{x}(\sqrt{x}-1)}`
`={\sqrt{x}(\sqrt{x}+1)}/{(\sqrt{x}-1)^2}:{x-1+\sqrt{x}+2-x}/{\sqrt{x}(\sqrt{x}-1)}`
`={\sqrt{x}(\sqrt{x}+1)}/{(\sqrt{x}-1)^2}:{\sqrt{x}+1}/{\sqrt{x}(\sqrt{x}-1)}`
`={\sqrt{x}(\sqrt{x}+1)}/{(\sqrt{x}-1)^2} . {\sqrt{x}(\sqrt{x}-1)}/{\sqrt{x}+1}`
`=x/{\sqrt{x}-1}`
Vậy `P=x/{\sqrt{x}-1}` với `x>0;x\ne 1`
