Đáp án:
`a.`Biểu thức `P` xác định khi $\begin{cases}x\ge0\\\sqrt{x}-1\ne0\\\sqrt{x}-x\ne0\end{cases}\\\Leftrightarrow \begin{cases}x\ge0\\x\ne1\\\sqrt{x}\ne0\\1-\sqrt{x}\ne0\end{cases}\\\Leftrightarrow \begin{cases}x\ge0\\x\ne1\\x\ne0\\x\ne1\end{cases}\\\Leftrightarrow \begin{cases}x>0\\x\ne1\end{cases}$
`b.`Rút gọn:
`P=(\frac{sqrtx}{sqrtx-1}+\frac{1}{sqrtx-x})xx \frac{sqrtx-1}{sqrtx+1}`
`= \frac{(sqrtx-x)sqrtx+sqrtx-1}{(sqrtx-1)(sqrtx-x)}xx\frac{-(1-sqrtx)}{sqrtx+1}`
`= \frac{x-xsqrtx+sqrtx-1}{(sqrtx-1)(sqrtx-x)}xx\frac{-(1-sqrtx)}{sqrtx+1}`
`= \frac{sqrtx(sqrtx-x)-(1-sqrtx)}{(sqrtx-1)(sqrtx-x)}xx\frac{-(1-sqrtx)}{sqrtx+1}`
`= \frac{sqrtxsqrtx(1-sqrtx)-(1-sqrtx)}{(sqrtx-1)(sqrtx-x)}xx\frac{-(1-sqrtx)}{sqrtx+1}`
`= \frac{(1-sqrtx)(sqrt{x xx x}-1)}{(sqrtx-1)(sqrtx-x)}xx\frac{-(1-sqrtx)}{sqrtx+1}`
`= \frac{-(sqrtx-1)(sqrt{x^2}-1)}{(sqrtx-1)(sqrtx-x)}xx\frac{-(1-sqrtx)}{sqrtx+1}`
`= \frac{-(|x|-1)}{sqrtx-x}xx\frac{-(1-sqrtx)}{sqrtx+1}`
`= \frac{-|x|+1}{sqrtx}xx\frac{-1}{sqrtx+1}`
`= \frac{-|x|+1}{sqrtx(sqrtx+1)}`
`= \frac{|x|-1}{sqrtx(sqrtx+1)}`
`= \frac{|x|-1}{x+sqrtx}`
`c.`Tìm `x`:
Để `P<0` thì `\frac{|x|-1}{x+sqrtx}<0` (`x>0, x\ne1`)
`<=> |x|-1<0`
`<=>`\(\left[ \begin{array}{l}x-1<0 \ (x\ge0)\\-x-1<0 \ (x<0)\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}x<1 \ (x\ge0)\\x>-1 \ (x<0)\end{array} \right.\)
`<=>`\(\left[ \begin{array}{l}0\le x<1\\-1<x<0\end{array} \right.\)
`<=> -1<x<1`
