a, ĐKXĐ:\(\left\{{}\begin{matrix}x-1\ne0\\1-x^2\ne0\\x+1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x^2\ne1\\x\ne-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)
\(A=\left(\dfrac{x}{x-1}+\dfrac{2x}{1-x^2}-\dfrac{1}{x+1}\right):\left(x-1\right)\)
\(\Rightarrow A=\left(\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{2x}{\left(1-x\right)\left(1+x\right)}-\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}\right).\dfrac{1}{x-1}\)
\(\Rightarrow A=\left(\dfrac{x^2+x}{\left(x-1\right)\left(x+1\right)}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}-\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}\right).\dfrac{1}{x-1}\)
\(\Rightarrow A=\dfrac{x^2+x-2x-x+1}{\left(x-1\right)\left(x+1\right)}.\dfrac{1}{x-1}\)
\(\Rightarrow A=\dfrac{x^2-2x+1}{\left(x-1\right)^2\left(x+1\right)}\)
\(\Rightarrow A=\dfrac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}\)
\(\Rightarrow A=\dfrac{1}{x+1}\)
b, Thay \(x=-\dfrac{1}{3}\) vào A ta được \(A=\dfrac{1}{x+1}=\dfrac{1}{-\dfrac{1}{3}+1}=\dfrac{1}{\dfrac{2}{3}}=\dfrac{3}{2}\)
