a/ ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ne0\\x+1\ne0\\x\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne-1\\x\ne0\end{matrix}\right.\)
Vậy ....
b/ Ta có :
\(B=\left(\frac{x^2+1}{x+1}-1\right)\left(\frac{4}{x-1}-\frac{2}{x}\right)\)
\(=\left(\frac{x^2+1}{x+1}-\frac{x+1}{x+1}\right)\left(\frac{4x}{x\left(x-1\right)}-\frac{2\left(x-1\right)}{x\left(x-1\right)}\right)\)
\(=\frac{x^2+1-x-1}{x+1}.\frac{4x-2x+2}{x\left(x-1\right)}\)
\(=\frac{x\left(x-1\right)}{x+1}.\frac{2\left(x+1\right)}{x\left(x-1\right)}\)
\(=2\)
Vậy...
