Сначала нужно перевести смешанные числа в неправильные дроби, затем привести дроби к общему знаменателю и выполнить сложение или вычитание.
\[1 \frac{3}{4} + 3 \frac{5}{8} = \frac{7}{4} + \frac{29}{8} = \frac{14}{8} + \frac{29}{8} = \frac{43}{8} = 5 \frac{3}{8}\]
\[\frac{23}{21} - \frac{5}{12} - \frac{1}{6} = \frac{23 \cdot 4}{21 \cdot 4} - \frac{5 \cdot 7}{12 \cdot 7} - \frac{1 \cdot 14}{6 \cdot 14} = \frac{92}{84} - \frac{35}{84} - \frac{14}{84} = \frac{92 - 35 - 14}{84} = \frac{43}{84}\]
\[\frac{5}{8} + 3 \frac{1}{12} + \frac{7}{12} = \frac{5}{8} + \frac{37}{12} + \frac{7}{12} = \frac{5 \cdot 3}{8 \cdot 3} + \frac{37 \cdot 2}{12 \cdot 2} + \frac{7 \cdot 2}{12 \cdot 2} = \frac{15}{24} + \frac{74}{24} + \frac{14}{24} = \frac{15 + 74 + 14}{24} = \frac{103}{24} = 4 \frac{7}{24}\]
\[\frac{5}{6} - \frac{1}{8} + \frac{5}{12} = \frac{5 \cdot 4}{6 \cdot 4} - \frac{1 \cdot 3}{8 \cdot 3} + \frac{5 \cdot 2}{12 \cdot 2} = \frac{20}{24} - \frac{3}{24} + \frac{10}{24} = \frac{20 - 3 + 10}{24} = \frac{27}{24} = \frac{9}{8} = 1 \frac{1}{8}\]
\[3 \frac{7}{14} + 4 \frac{9}{14} - 2 \frac{5}{21} = 3 \frac{1}{2} + 4 \frac{9}{14} - 2 \frac{5}{21} = \frac{7}{2} + \frac{65}{14} - \frac{47}{21} = \frac{7 \cdot 21}{2 \cdot 21} + \frac{65 \cdot 3}{14 \cdot 3} - \frac{47 \cdot 2}{21 \cdot 2} = \frac{147}{42} + \frac{195}{42} - \frac{94}{42} = \frac{147 + 195 - 94}{42} = \frac{248}{42} = \frac{124}{21} = 5 \frac{19}{21}\]
\[2 \frac{1}{3} + \frac{1}{5} - 1 \frac{1}{8} = \frac{7}{3} + \frac{1}{5} - \frac{9}{8} = \frac{7 \cdot 40}{3 \cdot 40} + \frac{1 \cdot 24}{5 \cdot 24} - \frac{9 \cdot 15}{8 \cdot 15} = \frac{280}{120} + \frac{24}{120} - \frac{135}{120} = \frac{280 + 24 - 135}{120} = \frac{169}{120} = 1 \frac{49}{120}\]