Đáp án:
$\begin{array}{l}
Dkxd:x \ne - 3;x \ne 3\\
A = \left( {\dfrac{{3 - x}}{{x + 3}}.\dfrac{{{x^2} + 6x + 9}}{{{x^2} - 9}} + \dfrac{x}{{x + 3}}} \right):\dfrac{{3{x^2}}}{{x + 3}}\\
= \left( {\dfrac{{ - \left( {x - 3} \right)}}{{x + 3}}.\dfrac{{{{\left( {x + 3} \right)}^2}}}{{\left( {x + 3} \right)\left( {x - 3} \right)}} + \dfrac{x}{{x + 3}}} \right).\dfrac{{x + 3}}{{3{x^2}}}\\
= \left( { - 1 + \dfrac{x}{{x + 3}}} \right).\dfrac{{x + 3}}{{3{x^2}}}\\
= \dfrac{{ - x - 3 + x}}{{x + 3}}.\dfrac{{x + 3}}{{3{x^2}}}\\
= \dfrac{{ - 3}}{{3{x^2}}}\\
= - \dfrac{1}{{{x^2}}}
\end{array}$
