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 \(x=-\dfrac{1}{2}\)
`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`
`<=>` $\left[ \begin{array}{l}x>0\\x<-3\end{array} \right.$
ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
a) Ta có: \(F=\left(\dfrac{x+1}{1-x}-\dfrac{1-x}{x+1}-\dfrac{4x^2}{x^2-1}\right):\dfrac{4x^2-4}{x^2-2x+1}\)
\(=\left(\dfrac{-\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}-\dfrac{4x^2}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{4\left(x^2-1\right)}{\left(x-1\right)^2}\)
\(=\dfrac{-x^2-2x-1+x^2-2x+1-4x^2}{\left(x-1\right)\left(x+1\right)}:\dfrac{4\left(x-1\right)\left(x+1\right)}{\left(x-1\right)^2}\)
\(=\dfrac{-4x^2-4x}{\left(x-1\right)\left(x+1\right)}:\dfrac{4\left(x+1\right)}{x-1}\)
\(=\dfrac{-4x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x-1}{4\left(x+1\right)}\)
\(=\dfrac{-4x}{x-1}\cdot\dfrac{x-1}{4\left(x+1\right)}\)
\(=\dfrac{-4x}{4\left(x+1\right)}\)
\(=\dfrac{-x}{x+1}\)
b) Để F<-1 thì F+1<0
\(\Leftrightarrow\dfrac{-x}{x+1}+1< 0\)
\(\Leftrightarrow\dfrac{-x+x+1}{x+1}< 0\)
\(\Leftrightarrow\dfrac{1}{x+1}< 0\)
mà 1>0
nên x+1<0
hay x<-1
Kết hợp ĐKXĐ, ta được: x<-1
giúp tớ với ạ. Cảm ơn mọi người rất nhiều ạ
