Люди пж не игнорте 6.70 тешите пж сам не могу

Решите задачу:

$-\frac{18x^2y^2}{5abm}:(-\frac{9xy^3}{5a^2m^4})=\frac{18x^2y^2\cdot 5a^2m^4}{5abm\cdot 9xy^3}=\frac{2xam^3}{by}\\\\35m^2n:\frac{7m^3}{24}=\frac{35m^2n\cdot \; 24}{7m^3}=\frac{120n}{m}\\\\\frac{13a}{12xy^2}\cdot 4x^2y=\frac{13ax}{3y}\\\\-\frac{18p^3}{11q^3}:\frac{9p^2}{22q^4}=-\frac{18p^3\; \cdot 22q^4}{11q^3\; \cdot 9p^2}=-\frac{2p\cdot 2q}{1}=-4pq$

$6ax^3:\frac{3x^2}{4a}=\frac{6ax^3\; \cdot 4a}{3x^2}=\frac{2ax\cdot 4a}{1}=8a^2x\\\\\frac{x^2y^2}{11ab^2}:\Big (-\frac{4xy^3}{33ab}\Big )=-\frac{x^2y^2\; \cdot 33ab}{11ab^2\; \cdot 4xy^3}=-\frac{x\cdot 3}{b\cdot 4y}=-\frac{3x}{4by}$