i) 6\frac{3}{4} - 2\frac{1}{2} j) 4\frac{3}{5} - 1\frac{1}{10} k) 6\frac{5}{7} - 5\frac{7}{10} l) 2\frac{3}{8} - 1\frac{7}{8} m) 1\frac{1}{4} - \frac{3}{1} n) 2\frac{1}{3} - 1\frac{1}{2} o) 7\frac{1}{9} - 4\frac{1}{3}

Ответ:

Решение:

i) \(6\frac{3}{4} - 2\frac{1}{2} = \frac{27}{4} - \frac{5}{2} = \frac{27}{4} - \frac{5 \cdot 2}{2 \cdot 2} = \frac{27}{4} - \frac{10}{4} = \frac{27-10}{4} = \frac{17}{4} = 4\frac{1}{4}\)

j) \(4\frac{3}{5} - 1\frac{1}{10} = \frac{23}{5} - \frac{11}{10} = \frac{23 \cdot 2}{5 \cdot 2} - \frac{11}{10} = \frac{46}{10} - \frac{11}{10} = \frac{46-11}{10} = \frac{35}{10} = 3\frac{5}{10} = 3\frac{1}{2}\)

k) \(6\frac{5}{7} - 5\frac{7}{10} = \frac{47}{7} - \frac{57}{10} = \frac{47 \cdot 10}{7 \cdot 10} - \frac{57 \cdot 7}{10 \cdot 7} = \frac{470}{70} - \frac{399}{70} = \frac{470-399}{70} = \frac{71}{70} = 1\frac{1}{70}\)

l) \(2\frac{3}{8} - 1\frac{7}{8} = \frac{19}{8} - \frac{15}{8} = \frac{19-15}{8} = \frac{4}{8} = \frac{1}{2}\)

m) Некорректные условия, так как вычитаемое больше уменьшаемого. \(1\frac{1}{4} - \frac{3}{1} = \frac{5}{4} - 3 = \frac{5}{4} - \frac{12}{4} = \frac{5-12}{4} = \frac{-7}{4} = -1\frac{3}{4}\)

n) \(2\frac{1}{3} - 1\frac{1}{2} = \frac{7}{3} - \frac{3}{2} = \frac{7 \cdot 2}{3 \cdot 2} - \frac{3 \cdot 3}{2 \cdot 3} = \frac{14}{6} - \frac{9}{6} = \frac{14-9}{6} = \frac{5}{6}\)

o) \(7\frac{1}{9} - 4\frac{1}{3} = \frac{64}{9} - \frac{13}{3} = \frac{64}{9} - \frac{13 \cdot 3}{3 \cdot 3} = \frac{64}{9} - \frac{39}{9} = \frac{64-39}{9} = \frac{25}{9} = 2\frac{7}{9}\)

Ответ:

i) \(4\frac{1}{4}\)

j) \(3\frac{1}{2}\)

k) \(1\frac{1}{70}\)

l) \(\frac{1}{2}\)

m) \(-1\frac{3}{4}\)

n) \(\frac{5}{6}\)

o) \(2\frac{7}{9}\)