Решение:
Раздел А
- \( \frac{3}{7} = \frac{3 \cdot 6}{7 \cdot 6} = \frac{18}{42} \)
- \( \frac{11}{9} = \frac{11 \cdot 11}{9 \cdot 11} = \frac{121}{99} \)
- \( \frac{2}{15} = \frac{2 \cdot 4}{15 \cdot 4} = \frac{8}{60} \)
- \( \frac{7}{12} = \frac{7 \cdot 4}{12 \cdot 4} = \frac{28}{48} \)
- \( \frac{a}{3} = \frac{a \cdot 5}{3 \cdot 5} = \frac{5a}{15} \)
- \( \frac{x}{9} = \frac{x \cdot 11}{9 \cdot 11} = \frac{11x}{99} \)
- \( \frac{2y}{3} = \frac{2y \cdot 13}{3 \cdot 13} = \frac{26y}{39} \)
- \( \frac{3y}{8} = \frac{3y \cdot 11}{8 \cdot 11} = \frac{33y}{88} \)
- \( \frac{3}{5} = \frac{3 \cdot 3}{5 \cdot 3} = \frac{9}{15} \)
- \( \frac{4}{x} = \frac{4 \cdot 3}{x \cdot 3} = \frac{12}{3x} \)
- \( \frac{2}{y} = \frac{2 \cdot 16}{y \cdot 16} = \frac{32}{16y} \)
- \( \frac{7}{b} = \frac{7 \cdot 4}{b \cdot 4} = \frac{28}{4b} \)
- \( \frac{1}{x^2} = \frac{1 \cdot 2}{x^2 \cdot 2} = \frac{2}{2x^2} \)
- \( \frac{5}{y^2} = \frac{5 \cdot 3}{y^2 \cdot 3} = \frac{15}{3y^2} \)
- \( \frac{7}{a^3} = \frac{7 \cdot 4}{a^3 \cdot 4} = \frac{28}{4a^3} \)
- \( \frac{2}{x^4} = \frac{2 \cdot 5}{x^4 \cdot 5} = \frac{10}{5x^4} \)
Раздел Б
- \( \frac{a}{b} = \frac{a \cdot a}{b \cdot a} = \frac{a^2}{ab} \)
- \( \frac{x}{y} = \frac{x \cdot y}{y \cdot y} = \frac{xy}{y^2} \)
- \( \frac{2}{x} = \frac{2 \cdot xy}{x \cdot xy} = \frac{2xy}{x^2y} \)
- \( \frac{7b}{a} = \frac{7b \cdot 3b}{a \cdot 3b} = \frac{21b^2}{3ab} \)
- \( \frac{a^2}{2x} = \frac{a^2 \cdot 4x}{2x \cdot 4x} = \frac{4a^2x}{8x^2} \)
- \( \frac{3x}{5b} = \frac{3x \cdot 4b}{5b \cdot 4b} = \frac{12xb}{20b^2} \)
- \( \frac{2a}{b^2} = \frac{2a \cdot ab}{b^2 \cdot ab} = \frac{2a^2b}{ab^3} \)
- \( \frac{1}{3ab} = \frac{1 \cdot 6b}{3ab \cdot 6b} = \frac{6b}{18ab^2} \)
- \( \frac{3}{x^2y^2} = \frac{3 \cdot 6y}{x^2y^2 \cdot 6y} = \frac{18y}{6x^2y^3} \)
Ответ: Заполненные равенства приведены в решении.