Разбираемся:
\[ \frac{5a}{a+3} \]
\[ \frac{a-6}{3a+9} = \frac{a-6}{3(a+3)} \]
\[ \frac{133}{6a-a^2} = \frac{133}{a(6-a)} = -\frac{133}{a(a-6)} \]
\[ \frac{5a}{a+3} + \frac{a-6}{3(a+3)} \cdot \left(-\frac{133}{a(a-6)}\right) \]
\[ \frac{5a}{a+3} - \frac{133}{3a(a+3)} \]
\[ \frac{5a \cdot 3a}{3a(a+3)} - \frac{133}{3a(a+3)} = \frac{15a^2 - 133}{3a(a+3)} \]
Ответ: \(\frac{15a^2 - 133}{3a(a+3)}\)