Đáp án:
$\begin{array}{l}
A = \left( {\frac{x}{{{x^2} - 9}} + \frac{3}{{3 - x}} + \frac{1}{{x + 3}}} \right):\left( {x - 3 + \frac{{10 - {x^2}}}{{x + 3}}} \right)\\
Dkxd:x \ne 3;x \ne - 3\\
A = \left( {\frac{x}{{\left( {x - 3} \right)\left( {x + 3} \right)}} - \frac{3}{{x - 3}} + \frac{1}{{x + 3}}} \right):\frac{{\left( {x - 3} \right)\left( {x + 3} \right) + 10 - {x^2}}}{{x + 3}}\\
= \frac{{x - 3\left( {x + 3} \right) + \left( {x - 3} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}.\frac{{x + 3}}{{{x^2} - 9 + 10 - {x^2}}}\\
= \frac{{x - 3x - 9 + x - 3}}{{x - 3}}.\frac{1}{1}\\
= \frac{{ - x - 12}}{{x - 3}}\\
A < 0\\
\Rightarrow \frac{{ - x - 12}}{{x - 3}} < 0\\
\Rightarrow \frac{{x + 12}}{{x - 3}} > 0\\
\Rightarrow \left[ \begin{array}{l}
x > 3\\
x < - 12
\end{array} \right.
\end{array}$
Vậy x>3 hoặc x<-12 thì A<0
