487. Прошу помочь. Внизу в фото все написано.

Решите задачу:

$\frac{4a^2b^2}{2a^4b^2} =\frac{2}{a^2}\; ;\; \frac{24p^4q^4}{48p^2q^2} = \frac{p^2q^2}{2} \; ;\; \frac{-ax^2}{xy} = \frac{-ax}{y} \; ;\; \frac{3xy^2}{6x^3y^3} = \frac{1}{2x^2y} \; ;\\\\ \frac{36nm^2}{18mn} = \frac{2m}{1} =2m\; ;\; \frac{-6ax}{-18ax} = \frac{1}{3} \; ;\; \frac{6ab^2}{9bc^2} = \frac{2ab}{3c^2} \; ;\; \frac{3axy}{6ay^3} = \frac{x}{2y^2} \; ;\\\\ \frac{-32b^5c}{12b^4c^2} = -\frac{8b}{3c} \; ;\; \frac{6xy-18x^2}{(y-3x)^3} = \frac{6x(y-3x)}{(y-3x)^3} = \frac{6x}{(y-3x)^2}$