Đáp án:
Giải thích các bước giải:
`P=(\frac{x-1}{x+1}-\frac{x+1}{x-1}):\frac{1}{x-1}`
ĐK: `x \ne \pm 1`
a) `P=[\frac{(x-1)^2}{(x-1)(x+1)}-\frac{(x+1)^2}{(x-1)(x+1)}].\frac{x-1}{1}`
`P=[\frac{x^2-2x+1}{(x-1)(x+1)}-\frac{x^2+2x+1}{(x-1)(x+1)}].\frac{x-1}{1}`
`P=[\frac{x^2-2x+1-x^2-2x-1}{(x-1)(x+1)}].\frac{x-1}{1}`
`P=[\frac{-4x}{(x-1)(x+1)}].\frac{x-1}{1}`
`P=\frac{-4x}{x+1}`
b) Thay `x=5` vào `P` ta có:
`P=\frac{-4.5}{5+1}`
`P=-10/3`
Vậy khi `x=5` thì `P=-10/3`
c) `P>1`
`⇔ \frac{-4x}{x+1} > 1`
`⇔ \frac{-4x}{x+1} - 1 > 0`
`⇔ \frac{-4x}{x+1} - \frac{x+1}{x+1} > 0`
`⇔ \frac{-4x-x-1}{x+1} > 0`
`⇔ \frac{-5x-1}{x+1} > 0`
`⇔` \(\left[ \begin{array}{l}\begin{cases} -5x-1>0\\x+1>0\end{cases}\\\begin{cases} -5x-1<0\\x+1<0\end{cases}\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}\begin{cases} x < -\dfrac{1}{5}\\x > -1\end{cases}\\\begin{cases} x > -\dfrac{1}{5}\\x < -1\end{cases}\end{array} \right.\)
`⇔ -1<x<-1/5`
Vậy ........
