Đáp án:
Giải thích các bước giải:
`P=(\frac{2}{\sqrt{x}-1}-\frac{2x+4}{x\sqrt{x}-1}):\frac{x-1}{x\sqrt{x}-1}` \((x \neq 1, x \geq 0\))
a) `P=(\frac{2.(x+\sqrt{x}+1)}{(\sqrt{x}-1).(x+\sqrt{x}+1)}-\frac{2x+4}{(\sqrt{x}-1).(x+\sqrt{x}+1))).\frac{(\sqrt{x}-1).(x+\sqrt{x}+1)}{x-1}`
`P=\frac{2x+2\sqrt{x}+2-2x-4}{(\sqrt{x}-1).(x+\sqrt{x}+1)}.\frac{(\sqrt{x}-1).(x+\sqrt{x}+1)}{x-1}`
`P=\frac{2\sqrt{x}-2}{(\sqrt{x}-1).(x+\sqrt{x}+1)}.\frac{(\sqrt{x}-1).(x+\sqrt{x}+1)}{(\sqrt{x}-1).(\sqrt{x}+1)}`
`P=\frac{2}{\sqrt{x}+1}`
b) `P>1`
`\Leftrightarrow\ \frac{2}{\sqrt{x}+1}>1`
`\Leftrightarrow\ \frac{2}{\sqrt{x}+1}-1>0`
`\Leftrightarrow\ \frac{2}{\sqrt{x}+1}-\frac{\sqrt{x}+1}{\sqrt{x}+1}>0`
`\Leftrightarrow\ \frac{1-\sqrt{x}}{\sqrt{x}+1}>0`
Ta có: `\sqrt{x}\geq0 \Rightarrow \sqrt{x}+1\geq1`
`\Rightarrow\ 1-\sqrt{x}>0`
`\Leftrightarrow\ x<1` kết hợp với ĐKXĐ
Vậy với `0\leq\ x< 1` thì `P>1`
