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