Тренажёр «Арифметические действия с обыкновенными дробями»

Решение:

Первый столбец:

1. \(\frac{1}{25} + \frac{43}{50} = \frac{2}{50} + \frac{43}{50} = \frac{45}{50} = \frac{9}{10}\)
2. \(\frac{1}{4} + \frac{37}{20} = \frac{5}{20} + \frac{37}{20} = \frac{42}{20} = \frac{21}{10}\)
3. \(\frac{1}{5} + \frac{17}{10} = \frac{2}{10} + \frac{17}{10} = \frac{19}{10}\)
4. \(\frac{1}{2} + \frac{11}{10} = \frac{5}{10} + \frac{11}{10} = \frac{16}{10} = \frac{8}{5}\)
5. \(\frac{1}{10} + \frac{29}{20} = \frac{2}{20} + \frac{29}{20} = \frac{31}{20}\)

Второй столбец:

1. \(\frac{4}{25} + \frac{15}{4} = \frac{16}{100} + \frac{375}{100} = \frac{391}{100}\)
2. \(\frac{9}{4} + \frac{8}{5} = \frac{45}{20} + \frac{32}{20} = \frac{77}{20}\)
3. \(\frac{14}{25} + \frac{3}{2} = \frac{28}{50} + \frac{75}{50} = \frac{103}{50}\)
4. \(\frac{11}{4} + \frac{6}{5} = \frac{55}{20} + \frac{24}{20} = \frac{79}{20}\)
5. \(\frac{3}{4} + \frac{7}{25} = \frac{75}{100} + \frac{28}{100} = \frac{103}{100}\)

Третий столбец:

1. \(\frac{3}{5} \cdot \frac{3}{2} = \frac{9}{10}\)
2. \(\frac{3}{5} \cdot \frac{2}{1} = \frac{6}{5}\)
3. \(\frac{9}{5} \cdot \frac{1}{20} = \frac{9}{100}\)
4. \(\frac{15}{2} \cdot \frac{7}{4} = \frac{105}{8}\)
5. \(\frac{15}{4} \cdot \frac{3}{2} = \frac{45}{8}\)

Четвёртый столбец:

1. \(\frac{5}{4} \cdot 8 = \frac{5 \cdot 8}{4} = 10\)
2. \(\frac{9}{2} \cdot 7 = \frac{63}{2}\)
3. \(\frac{3}{20} \cdot 5 = \frac{15}{20} = \frac{3}{4}\)
4. \(\frac{17}{4} \cdot 12 = \frac{17 \cdot 12}{4} = 17 \cdot 3 = 51\)
5. \(\frac{13}{5} \cdot 13 = \frac{169}{5}\)

Пятый столбец:

1. \(5: \frac{5}{15} = 5 \cdot \frac{15}{5} = 15\)
2. \(\frac{12}{5} : 4 = \frac{12}{5 \cdot 4} = \frac{3}{5}\)
3. \(\frac{3}{4} : 4 = \frac{3}{4 \cdot 4} = \frac{3}{16}\)
4. \(\frac{9}{2} : 25 = \frac{9}{2 \cdot 25} = \frac{9}{50}\)
5. \(\frac{1}{10} : 2 = \frac{1}{10 \cdot 2} = \frac{1}{20}\)

Шестой столбец:

1. \(\frac{3}{4} - \frac{1}{6} = \frac{9}{12} - \frac{2}{12} = \frac{7}{12} \Rightarrow \frac{7}{12} \cdot 3 = \frac{7}{4}\)
2. \(\frac{1}{6} + \frac{1}{4} = \frac{2}{12} + \frac{3}{12} = \frac{5}{12} \Rightarrow \frac{5}{12} \cdot 9 = \frac{5 \cdot 3}{4} = \frac{15}{4}\)
3. \(\frac{2}{5} + \frac{13}{15} = \frac{6}{15} + \frac{13}{15} = \frac{19}{15} \Rightarrow \frac{19}{15} \cdot 6 = \frac{19 \cdot 2}{5} = \frac{38}{5}\)
4. \(\frac{1}{5} + \frac{8}{15} = \frac{3}{15} + \frac{8}{15} = \frac{11}{15} \Rightarrow \frac{11}{15} \cdot 6 = \frac{11 \cdot 2}{5} = \frac{22}{5}\)
5. \(\frac{9}{10} - \frac{7}{15} = \frac{27}{30} - \frac{14}{30} = \frac{13}{30} \Rightarrow \frac{13}{30} \cdot 3 = \frac{13}{10}\)

Седьмой столбец:

1. \(\frac{27}{20} - \frac{1}{5} = \frac{27}{20} - \frac{4}{20} = \frac{23}{20}\)
2. \(\frac{13}{2} - \frac{12}{50} = \frac{13 \cdot 25}{50} - \frac{12}{50} = \frac{325-12}{50} = \frac{313}{50}\)
3. \(\frac{39}{50} - \frac{1}{10} = \frac{39}{50} - \frac{5}{50} = \frac{34}{50} = \frac{17}{25}\)
4. \(\frac{7}{50} - \frac{1}{25} = \frac{7}{50} - \frac{2}{50} = \frac{5}{50} = \frac{1}{10}\)
5. \(\frac{41}{50} - \frac{11}{5} = \frac{41}{50} - \frac{110}{50} = \frac{-69}{50}\)

Восьмой столбец:

1. \(\frac{1}{2} - \frac{9}{25} = \frac{25}{50} - \frac{18}{50} = \frac{7}{50}\)
2. \(\frac{11}{4} - \frac{2}{5} = \frac{55}{20} - \frac{8}{20} = \frac{47}{20}\)
3. \(\frac{3}{4} - \frac{1}{5} = \frac{15}{20} - \frac{4}{20} = \frac{11}{20}\)
4. \(\frac{1}{4} - \frac{3}{25} = \frac{25}{100} - \frac{12}{100} = \frac{13}{100}\)
5. \(\frac{17}{5} - \frac{13}{4} = \frac{68}{20} - \frac{65}{20} = \frac{3}{20}\)