ĐKXĐ : \(\left\{{}\begin{matrix}1-x\ne0\\x+1\ne0\\1-2x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne-1\\x\ne\dfrac{1}{2}\end{matrix}\right.\)
\(a,A=\left[\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right]:\dfrac{1-2x}{x^2-1}\)
\(\Leftrightarrow\left(\dfrac{-1\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}+\dfrac{2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{5-x}{\left(x+1\right)\left(x-1\right)}\right).\dfrac{1-2x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x-1+2x-2-5+x}{\left(x+1\right)\left(x-1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(=\dfrac{2x-8}{1-2x}\)
b, Để \(A>0\)
\(\Rightarrow\dfrac{2x-8}{1-2x}>0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-8>0\\1-2x>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-8< 0\\1-2x< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>4\\x< \dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 4\\x>\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
TH1 : Pt vô nghiệm
Vậy \(\left\{{}\begin{matrix}x< 4\\x>\dfrac{1}{2}\end{matrix}\right.\) thì A > 0
