Đáp án:
$\frac{{{x^2} + 3}}{{2{{(x - 3)}^2}}}$
Giải thích các bước giải:ĐKXĐ
$\{ _{x \ne 3}^{x \ne \pm 1}$
$\begin{array}{l}
\left( {\frac{2}{{x - 1}} - \frac{1}{{x + 1}}} \right)\left( {\frac{{{x^2} - 1}}{{{x^2} - 6x + 9}}} \right) + \frac{{x + 1}}{{2x - 6}}\\
= \frac{{2(x + 1) - (x - 1)}}{{(x - 1)(x + 1)}}.\frac{{{x^2} - 1}}{{{{(x - 3)}^2}}} + \frac{{x + 1}}{{2(x - 3)}}\\
= \frac{{x + 3}}{{{x^2} - 1}}.\frac{{{x^2} - 1}}{{{{(x - 3)}^2}}} + \frac{{(x + 1)(x - 3)}}{{2{{(x - 3)}^2}}}\\
= \frac{{2(x + 3)}}{{2{{(x - 3)}^2}}} + \frac{{{x^2} - 2x - 3}}{{2{{(x - 3)}^2}}}\\
= \frac{{{x^2} + 3}}{{2{{(x - 3)}^2}}}
\end{array}$
