Для решения данного задания необходимо сократить дроби, разделив числитель и знаменатель на их наибольший общий делитель (НОД).
1) \(\frac{42}{24}, \frac{35}{77}, \frac{48}{60}, \frac{72}{96}\)
- \(\frac{42}{24} = \frac{7 \cdot 6}{4 \cdot 6} = \frac{7}{4}\)
- \(\frac{35}{77} = \frac{5 \cdot 7}{11 \cdot 7} = \frac{5}{11}\)
- \(\frac{48}{60} = \frac{4 \cdot 12}{5 \cdot 12} = \frac{4}{5}\)
- \(\frac{72}{96} = \frac{3 \cdot 24}{4 \cdot 24} = \frac{3}{4}\)
2) \(\frac{10}{100}, \frac{400}{1000}, \frac{600}{800}, \frac{800}{1000}\)
- \(\frac{10}{100} = \frac{1 \cdot 10}{10 \cdot 10} = \frac{1}{10}\)
- \(\frac{400}{1000} = \frac{2 \cdot 200}{5 \cdot 200} = \frac{2}{5}\)
- \(\frac{600}{800} = \frac{3 \cdot 200}{4 \cdot 200} = \frac{3}{4}\)
- \(\frac{800}{1000} = \frac{4 \cdot 200}{5 \cdot 200} = \frac{4}{5}\)
3) \(\frac{72}{90}, \frac{60}{105}, \frac{45}{150}, \frac{84}{120}\)
- \(\frac{72}{90} = \frac{4 \cdot 18}{5 \cdot 18} = \frac{4}{5}\)
- \(\frac{60}{105} = \frac{4 \cdot 15}{7 \cdot 15} = \frac{4}{7}\)
- \(\frac{45}{150} = \frac{3 \cdot 15}{10 \cdot 15} = \frac{3}{10}\)
- \(\frac{84}{120} = \frac{7 \cdot 12}{10 \cdot 12} = \frac{7}{10}\)
4) \(\frac{56}{70}, \frac{36}{90}, \frac{66}{110}, \frac{96}{160}\)
- \(\frac{56}{70} = \frac{4 \cdot 14}{5 \cdot 14} = \frac{4}{5}\)
- \(\frac{36}{90} = \frac{2 \cdot 18}{5 \cdot 18} = \frac{2}{5}\)
- \(\frac{66}{110} = \frac{3 \cdot 22}{5 \cdot 22} = \frac{3}{5}\)
- \(\frac{96}{160} = \frac{3 \cdot 32}{5 \cdot 32} = \frac{3}{5}\)
Ответ: См. выше