Question

[Ngọc ánh]

giúp tớ với ạ, cảm ơn trước

Answer 1

[Nguyễn Lê Phước Thịnh]

c) Ta có: \(P=2x+\dfrac{1}{x+1}\)

\(\Leftrightarrow\dfrac{-x}{x+1}=2x+\dfrac{1}{x+1}\)

\(\Leftrightarrow\dfrac{-x}{x+1}=\dfrac{2x\left(x+1\right)+1}{x+1}\)

Suy ra: \(2x^2+2x+1=-x\)

\(\Leftrightarrow2x^2+3x+1=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\left(loại\right)\\x=-\dfrac{1}{2}\left(nhận\right)\end{matrix}\right.\)

Vậy: Để \(P=2x+\dfrac{1}{x+1}\) thì \(x=-\dfrac{1}{2}\)

Answer 2

[Yeutoanhoc]

`a)F=((x+1)/(1-x)-(1-x)/(x+1)-(4x^2)/(x^2-1)):(4x^2-4)/(x^2-2x+1)`

`đk:x ne +-1`

`F=((-(x+1)^2+(x-1)^2-4x^2)/(x^2-1)):(4(x-1)(x+1))/(x-1)^2`

`=(-x^2-2x-1+x^2-2x+1-4x^2)/(x^2-1):(4(x+1))/(x-1)`

`=(-4x^2-4x)/((x-1)(x+1)).(x-1)/(4(x+1))`

`=(-4(x-1))/((x-1)(x+1)).(x-1)/(4(x+1))`

`=-4/(x+1).(x-1)/(4(x+1)`

`=(1-x)/(x+1)^2`

`F<-1`

`<=>(1-x-(x+1)^2)/(x+1)^2<0`

Vì `(x+1)^2>0`

`=>1-x-(x+1)^2<0`

`<=>(x+1)^2+x-1>0`

`<=>x^2+2x+1+x-1>0`

`<=>x^2+3x>0`

`<=>x(x+3)>0`

`<=>` $\left[ \begin{array}{l}x>0\\x<-3\end{array} \right.$