Решение:
- a) 5ab-5 ⋅ 0,4a-3b8 = (5 ⋅ 0,4) ⋅ (a ⋅ a-3) ⋅ (b-5 ⋅ b8) = 2 ⋅ a1+(-3) ⋅ b-5+8 = 2a-2b3 = \(\frac{2b^3}{a^2}\).
- б) (0,3p6q-1)-2 = (0,3)-2 ⋅ (p6)-2 ⋅ (q-1)-2 = \(\left(\frac{3}{10}\right)^{-2}\) ⋅ p6⋅(-2) ⋅ q(-1)⋅(-2) = \(\left(\frac{10}{3}\right)^{2}\) ⋅ p-12 ⋅ q2 = \(\frac{100}{9}\) ⋅ \(\frac{1}{p^{12}}\)⋅q2 = \(\frac{100q^2}{9p^{12}}\).
Ответ: а) \(\frac{2b^3}{a^2}\), б) \(\frac{100q^2}{9p^{12}}\)