487. 1) (-\(\frac{2}{3}\)a³x²y³): (-\(\frac{1}{3}\)a²xy²); 2) (-\(\frac{3}{4}\)a²b³c): (1\(\frac{1}{2}\)a¹b¹c).

Решение:

1. \( \frac{-\frac{2}{3}a^3x^2y^3}{-\frac{1}{3}a^2xy^2} = \frac{2}{3} \cdot \frac{3}{1} a^{3-2}x^{2-1}y^{3-2} = 2ax y \)
2. \( \frac{-\frac{3}{4}a^2b^3c}{1\frac{1}{2}a b c} = \frac{-\frac{3}{4}a^2b^3c}{\frac{3}{2}a b c} = -\frac{3}{4} \cdot \frac{2}{3} a^{2-1}b^{3-1}c^{1-1} = -\frac{1}{2}ab^2 \)

Ответ: 1) 2axy; 2) -\(\frac{1}{2}\)ab².