Решение:
73. Умножение:
- \( \frac{a}{5} \cdot \frac{2b}{3} = \frac{a \cdot 2b}{5 \cdot 3} = \frac{2ab}{15} \)
- \( \frac{3}{m} \cdot \frac{3}{n} = \frac{3 \cdot 3}{m \cdot n} = \frac{9}{mn} \)
- \( \frac{2a^2}{x} \cdot \frac{a^2}{1} = \frac{2a^2 \cdot a^2}{x \cdot 1} = \frac{2a^4}{x} \)
- \( \frac{1}{a-1} \cdot \frac{2}{a-1} = \frac{1 \cdot 2}{(a-1)(a-1)} = \frac{2}{(a-1)^2} \)
74. Деление:
- \( \frac{1}{a} : \frac{1}{d} = \frac{1}{a} \cdot \frac{d}{1} = \frac{d}{a} \)
- \( \frac{n}{m^2} : \frac{m}{n} = \frac{n}{m^2} \cdot \frac{n}{m} = \frac{n \cdot n}{m^2 \cdot m} = \frac{n^2}{m^3} \)
- \( \frac{n+3}{2} : \frac{1}{n+3} = \frac{n+3}{2} \cdot \frac{n+3}{1} = \frac{(n+3)^2}{2} \)
- \( \frac{1}{b+5} : \frac{b-5}{b} = \frac{1}{b+5} \cdot \frac{b}{b-5} = \frac{b}{(b+5)(b-5)} = \frac{b}{b^2-25} \)
75. Действия:
- \( \frac{a}{bc} : \frac{c^3}{3a} = \frac{a}{bc} \cdot \frac{3a}{c^3} = \frac{a \cdot 3a}{bc \cdot c^3} = \frac{3a^2}{bc^4} \)
- \( \frac{x^2}{y} : \frac{x}{2y} = \frac{x^2}{y} \cdot \frac{2y}{x} = \frac{x^2 \cdot 2y}{y \cdot x} = 2x \)
- \( \frac{a^4b^2}{5xy} : \frac{10x^3}{a^2b^2} = \frac{a^4b^2}{5xy} \cdot \frac{a^2b^2}{10x^3} = \frac{a^4b^2 \cdot a^2b^2}{5xy \cdot 10x^3} = \frac{a^6b^4}{50x^4y} \)
- \( \frac{a^4}{b^3} : \frac{a^3}{b^2} = \frac{a^4}{b^3} \cdot \frac{b^2}{a^3} = \frac{a^4 \cdot b^2}{b^3 \cdot a^3} = \frac{a}{b} \)
- \( \frac{3mn}{2pq} \cdot \frac{6m^2}{pq} = \frac{3mn \cdot 6m^2}{2pq \cdot pq} = \frac{18m^3n}{2p^2q^2} = \frac{9m^3n}{p^2q^2} \)
Ответ: 73. а) 2ab/15, б) 9/mn, в) 2a^4/x, г) 2/(a-1)^2. 74. а) d/a, б) n^2/m^3, в) (n+3)^2/2, г) b/(b^2-25). 75. а) 3a^2/bc^4, б) 2x, в) a^6b^4/50x^4y, г) a/b, д) 9m^3n/p^2q^2.