4) 4a2 - 24a +36 a³ +1; 7a²-7a+7 Sa²-45

4) $$\frac{4a^2 - 24a + 36}{7a^2 - 7a + 7} \cdot \frac{a^3 + 1}{5a^2 - 45} = \frac{4(a^2 - 6a + 9)}{7(a^2 - a + 1)} \cdot \frac{(a+1)(a^2 - a + 1)}{5(a^2 - 9)} = \frac{4(a-3)^2 \cdot (a+1)(a^2 - a + 1)}{7(a^2 - a + 1) \cdot 5(a-3)(a+3)} = \frac{4(a-3)(a+1)}{35(a+3)}$$

Ответ: $$\frac{4(a-3)(a+1)}{35(a+3)}$$