Решение:
Для решения данного задания необходимо перемножить дроби.
- \( \frac{6}{11} \cdot \frac{4}{7} = \frac{6 \cdot 4}{11 \cdot 7} = \frac{24}{77} \)
- \( \frac{7}{20} \cdot \frac{10}{21} = \frac{7 \cdot 10}{20 \cdot 21} = \frac{70}{420} = \frac{7}{42} = \frac{1}{6} \)
- \( \frac{8}{9} \cdot \frac{27}{32} = \frac{8 \cdot 27}{9 \cdot 32} = \frac{8 \cdot 3 \cdot 9}{9 \cdot 4 \cdot 8} = \frac{3}{4} \)
- \( \frac{23}{28} \cdot \frac{49}{46} = \frac{23 \cdot 49}{28 \cdot 46} = \frac{23 \cdot 7 \cdot 7}{4 \cdot 7 \cdot 2 \cdot 23} = \frac{7}{8} \)
- \( \frac{34}{86} \cdot \frac{43}{51} = \frac{34 \cdot 43}{86 \cdot 51} = \frac{2 \cdot 17 \cdot 43}{2 \cdot 43 \cdot 3 \cdot 17} = \frac{1}{3} \)
- \( \frac{7}{18} \cdot \frac{90}{77} = \frac{7 \cdot 90}{18 \cdot 77} = \frac{7 \cdot 5 \cdot 18}{18 \cdot 7 \cdot 11} = \frac{5}{11} \)
- \( \frac{63}{64} \cdot \frac{48}{91} = \frac{63 \cdot 48}{64 \cdot 91} = \frac{9 \cdot 7 \cdot 3 \cdot 16}{4 \cdot 16 \cdot 7 \cdot 13} = \frac{27}{52} \)
- \( \frac{19}{100} \cdot \frac{5}{38} = \frac{19 \cdot 5}{100 \cdot 38} = \frac{19 \cdot 5}{20 \cdot 5 \cdot 2 \cdot 19} = \frac{1}{40} \)
Ответ: 1) \( \frac{24}{77} \); 2) \( \frac{1}{6} \); 3) \( \frac{3}{4} \); 4) \( \frac{7}{8} \); 5) \( \frac{1}{3} \); 6) \( \frac{5}{11} \); 7) \( \frac{27}{52} \); 8) \( \frac{1}{40} \).