Решение задания 1

a) $$\frac{42x^5}{y^4} \cdot \frac{y^2}{14x^5} = \frac{42}{14} \cdot \frac{x^5}{x^5} \cdot \frac{y^2}{y^4} = 3 \cdot 1 \cdot \frac{1}{y^2} = \frac{3}{y^2}$$.

б) $$\frac{63a^3b}{c} : (18a^2b) = \frac{63a^3b}{c} \cdot \frac{1}{18a^2b} = \frac{63}{18} \cdot \frac{a^3}{a^2} \cdot \frac{b}{b} \cdot \frac{1}{c} = \frac{7}{2} \cdot a \cdot 1 \cdot \frac{1}{c} = \frac{7a}{2c}$$.

в) $$\frac{4a^2-1}{a^2-9} \cdot \frac{6a+3}{a+3} = \frac{(2a-1)(2a+1)}{(a-3)(a+3)} \cdot \frac{3(2a+1)}{a+3} = \frac{3(2a-1)(2a+1)^2}{(a-3)(a+3)^2}$$.

г) $$\frac{p-q}{p} \cdot (\frac{p}{p-q} + \frac{p}{q}) = \frac{p-q}{p} \cdot \frac{pq + p(p-q)}{(p-q)q} = \frac{p-q}{p} \cdot \frac{pq + p^2 - pq}{(p-q)q} = \frac{p-q}{p} \cdot \frac{p^2}{(p-q)q} = \frac{p}{q}$$.