Đáp án:
\[\frac{{x - 3}}{{2\left( {x + 3} \right)}}\]
Giải thích các bước giải:
ĐKXĐ:\(\left\{ \begin{array}{l}
x - 3 \ne 0\\
x + 3 \ne 0
\end{array} \right. \Leftrightarrow x \ne \pm 3\)
Ta có:
\(\begin{array}{l}
\left( {\frac{1}{{x - 3}}.\frac{1}{{x + 3}}} \right):\frac{2}{{{x^2} - 6x + 9}}\\
= \frac{1}{{\left( {x - 3} \right)\left( {x + 3} \right)}}:\frac{2}{{{{\left( {x - 3} \right)}^2}}}\\
= \frac{{{{\left( {x - 3} \right)}^2}}}{{2\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \frac{{x - 3}}{{2\left( {x + 3} \right)}}
\end{array}\)
