ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ne0\\x+1\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)
Vậy...
b/ \(A=\left(\frac{2x+1}{x-1}+\frac{8}{x^2-1}-\frac{x-1}{x+1}\right).\frac{x^2-1}{5}\)
\(=\left(\frac{2x+1}{x-1}+\frac{8}{\left(x-1\right)\left(x+1\right)}-\frac{x-1}{x+1}\right).\frac{\left(x-1\right)\left(x+1\right)}{5}\)
\(=\left(\frac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{8}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\right).\frac{\left(x-1\right)\left(x+1\right)}{5}\)
\(=\frac{2x^2+2x+x+1-8-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)}{5}\)
\(=\frac{x^2+5x-8}{\left(x-1\right)\left(x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)}{5}\)
\(=\frac{x^2+5x-8}{5}\)
Vậy..
