Решение:

1. a) $$\frac{36a^2b^3}{18a^5b} = \frac{36}{18} \cdot \frac{a^2}{a^5} \cdot \frac{b^3}{b} = 2 \cdot \frac{1}{a^3} \cdot b^2 = \frac{2b^2}{a^3}$$
2. б) $$\frac{72y^3b}{c}:(27y^2b) = \frac{72y^3b}{c} \cdot \frac{1}{27y^2b} = \frac{72}{27} \cdot \frac{y^3}{y^2} \cdot \frac{b}{b} \cdot \frac{1}{c} = \frac{8}{3} \cdot y \cdot 1 \cdot \frac{1}{c} = \frac{8y}{3c}$$
3. в) $$\frac{9a^2-1}{a^2-25} \cdot \frac{6a-2}{a+5} = \frac{(3a-1)(3a+1)}{(a-5)(a+5)} \cdot \frac{2(3a-1)}{a+5} = \frac{2(3a-1)^2(3a+1)}{(a-5)(a+5)^2}$$
4. г) $$\frac{p-q}{p}-(\frac{p}{p-q}+\frac{q}{q-p}) = \frac{p-q}{p} - (\frac{p}{p-q} - \frac{q}{p-q}) = \frac{p-q}{p} - \frac{p-q}{p-q} = \frac{p-q}{p} - 1 = \frac{p-q}{p} - \frac{p}{p} = \frac{p-q-p}{p} = \frac{-q}{p} = -\frac{q}{p}$$

Ответ:

• a) $$\frac{2b^2}{a^3}$$
• б) $$\frac{8y}{3c}$$
• в) $$\frac{2(3a-1)^2(3a+1)}{(a-5)(a+5)^2}$$
• г) $$\frac{-q}{p}$$