Đáp án:
$\begin{array}{l}
a)Dkxd:\left\{ \begin{array}{l}
x - 1 \ne 0\\
{x^2} - 1 \ne 0\\
x + 1 \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ne 1\\
x \ne - 1
\end{array} \right.\\
b)B = \left( {\frac{{2x + 1}}{{x - 1}} + \frac{8}{{{x^2} - 1}} - \frac{{x - 1}}{{x + 1}}} \right).\frac{{{x^2} - 1}}{5}\\
= \frac{{\left( {2x + 1} \right)\left( {x + 1} \right) + 8 - \left( {x - 1} \right)\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}.\frac{{\left( {x + 1} \right)\left( {x - 1} \right)}}{5}\\
= \frac{{2{x^2} + 3x + 1 + 8 - {x^2} + 2x - 1}}{5}\\
= \frac{{{x^2} + 5x + 8}}{5}\\
= \frac{{{{\left( {x + \frac{5}{2}} \right)}^2} + \frac{7}{4}}}{5} > 0\forall x\\
Vậy\,B > 0\forall x
\end{array}$
