Решение:
- \( \frac{8a^2b}{2ab} = 4a^{2-1}b^{1-1} = 4a \cdot 1 = 4a \)
- \( \frac{-9a^3b^2}{3ab} = -3a^{3-1}b^{2-1} = -3a^2b \)
- \( \frac{16x^2y^3}{4x^2y} = 4x^{2-2}y^{3-1} = 4y^2 \)
- \( \frac{20m^4n^5}{5m^2n^3} = 4m^{4-2}n^{5-3} = 4m^2n^2 \)
Ответ: 1) 4a; 2) -3a²b; 3) 4y²; 4) 4m²n².