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Решите задачу:

$g) \frac{6a+12b}{a^2-2ab+b^2}: \frac{4a+8b}{a^2-b^2}= \frac{6(a+b)^2(a-b)}{4(a+b)(a-b)^2}= \frac{3(a+b)}{2(a-b)} \\ d) \frac{p^2+pq+q^2}{pq+3q^2}: \frac{p^3-q^3}{p^2-3pq}= \frac{p(p-3q)(p^2+pq+q^2)}{q(p+3q)(p-q)(p^2+pq+q^2)}= \frac{p(p-3q)}{q(p+3q)(p-q)} \\ e) \frac{a^3-8b^3}{a^2-3ab}: \frac{3a^2+6ab+12b^2}{3b^2-ab}= \frac{b(a-2b)(a^2+2ab+4b^2)(3b-a)}{3a(a-3b)(a^2+2ab+4b^2)}= \frac{b(2b-a)}{3a}$