\(a,ĐKXĐ:\hept{\begin{cases}x-1\ne0\\x+1\ne0\end{cases}\Leftrightarrow x\ne\pm1}\)
\(b,A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\left(\frac{1}{x+1}+\frac{x}{1-x}+\frac{2}{x^2-1}\right)\)
\(=\frac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}:\frac{x-1-x\left(x+1\right)+2}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)}{x-1-x^2-x+2}\)
\(=\frac{4x}{1-x^2}\)
\(c,A\ge0\Leftrightarrow\frac{4x}{1-x^2}\ge0\)
\(\Leftrightarrow\hept{\begin{cases}4x\ge0\\1-x^2\ge0\end{cases}\left(h\right)\hept{\begin{cases}4x\le0\\1-x^2\le0\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}x\ge0\\x^2\le1\end{cases}\left(h\right)\hept{\begin{cases}x\le0\\x^2\ge1\end{cases}}}\)
\(\Leftrightarrow0\le x\le1\left(h\right)x\le-1\)
Vậy ///////