(a - \frac{9a-9}{a+3}) : \frac{a^2-3a}{a+3}

Преобразуем выражение:

1. \(a - \frac{9a-9}{a+3} = \frac{a(a+3)-(9a-9)}{a+3} = \frac{a^2+3a-9a+9}{a+3} = \frac{a^2-6a+9}{a+3}\)
2. \(\frac{a^2-6a+9}{a+3} : \frac{a^2-3a}{a+3} = \frac{a^2-6a+9}{a+3} \cdot \frac{a+3}{a^2-3a} = \frac{(a-3)^2}{a(a-3)} = \frac{a-3}{a}\)

Ответ: \(\frac{a-3}{a}\)