Đáp án:
$\begin{array}{l}
a)Đkxđ:\left\{ \begin{array}{l}
x - 1 \ne 0\\
x + 1 \ne 0\\
{x^2} - 1 \ne 0
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x \ne 1\\
x \ne - 1
\end{array} \right.\\
b)\\
A = \left( {\frac{1}{{x - 1}} - \frac{1}{{x + 1}} + 1} \right):\frac{1}{{{x^2} - 1}}\\
= \frac{{x + 1 - \left( {x - 1} \right) + \left( {x - 1} \right)\left( {x + 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}.\left( {x + 1} \right).\left( {x - 1} \right)\\
= x + 1 - x + 1 + {x^2} - 1\\
= {x^2} + 1\\
Thay\,x = 3\sqrt 2 \left( {tmdk} \right)vào\,A\\
\Rightarrow A = {\left( {3\sqrt 2 } \right)^2} + 1 = 18 + 1 = 19
\end{array}$