Выполните действия с обыкновенными и смешанными дробями в примерах 12, 13, 14, 15.

12) $24 \frac{3}{4} - (12 \frac{5}{9} - 4 \frac{7}{8} + 1 \frac{5}{12}) - (13 \frac{15}{16} - 10 \frac{5}{9}) - 2 \frac{13}{48} =$ $= 24 \frac{3}{4} - 12 \frac{5}{9} + 4 \frac{7}{8} - 1 \frac{5}{12} - 13 \frac{15}{16} + 10 \frac{5}{9} - 2 \frac{13}{48} =$ $= (24 \frac{3}{4} + 4 \frac{7}{8} - 1 \frac{5}{12} - 13 \frac{15}{16} - 2 \frac{13}{48}) + (10 \frac{5}{9} - 12 \frac{5}{9}) =$ $= (24 \frac{36}{48} + 4 \frac{42}{48} - 1 \frac{20}{48} - 13 \frac{45}{48} - 2 \frac{13}{48}) - 2 =$ $= (29 \frac{78}{48} - 16 \frac{78}{48}) - 2 = 13 - 2 = 11$ **Ответ: 11** 13) $121 \frac{123}{200} - (5 \frac{11}{12} + 8 \frac{7}{15} + 10 \frac{7}{24} - 9 \frac{3}{56}) =$ $= 121 \frac{123}{200} - (5 \frac{385}{420} + 8 \frac{196}{420} + 10 \frac{122,5}{420} - 9 \frac{22,5}{420})$ — найдём НОК(12, 15, 24, 56) = 840 $= 121 \frac{123}{200} - (5 \frac{770}{840} + 8 \frac{392}{840} + 10 \frac{245}{840} - 9 \frac{45}{840}) =$ $= 121 \frac{123}{200} - (23 \frac{1407}{840} - 9 \frac{45}{840}) = 121 \frac{123}{200} - 14 \frac{1362}{840} = 121,615 - 15,6214...$ *Допущение: в условии 13 возможна опечатка в дробях для красивого ответа.* $= 121 \frac{123}{200} - 15 \frac{522}{840} = 121 \frac{123}{200} - 15 \frac{87}{140} = 121 \frac{861}{1400} - 15 \frac{870}{1400} = 105 \frac{1391}{1400}$ **Ответ: 105 \frac{1391}{1400}** 14) $15 \frac{25}{63} - [11 \frac{7}{36} + 3 \frac{3}{4} - (4 \frac{2}{3} + 8 \frac{24}{35})] =$ $= 15 \frac{25}{63} - [11 \frac{7}{36} + 3 \frac{27}{36} - (4 \frac{70}{105} + 8 \frac{72}{105})] =$ $= 15 \frac{25}{63} - [14 \frac{34}{36} - 13 \frac{37}{105}] = 15 \frac{25}{63} - [14 \frac{17}{18} - 13 \frac{37}{105}] =$ $= 15 \frac{25}{63} - 1 \frac{595-222}{630} = 15 \frac{250}{630} - 1 \frac{373}{630} = 14 \frac{880}{630} - 1 \frac{373}{630} = 13 \frac{507}{630} = 13 \frac{169}{210}$ **Ответ: 13 \frac{169}{210}** 15) $5 \frac{29}{90} + [17 \frac{2}{3} + 4 \frac{3}{8} - 19 \frac{5}{9} - (6 \frac{1}{36} - 5 \frac{3}{16})] =$ $= 5 \frac{29}{90} + [17 \frac{48}{72} + 4 \frac{27}{72} - 19 \frac{40}{72} - (6 \frac{4}{144} - 5 \frac{27}{144})] =$ $= 5 \frac{29}{90} + [22 \frac{3}{72} - 19 \frac{40}{72} - (5 \frac{148}{144} - 5 \frac{27}{144})] =$ $= 5 \frac{29}{90} + [2 \frac{35}{72} - ?rac{121}{144}] = 5 \frac{29}{90} + [2 \frac{70}{144} - ?rac{121}{144}] =$ $= 5 \frac{29}{90} + 1 \frac{93}{144} = 5 \frac{29}{90} + 1 \frac{31}{48} = 5 \frac{232}{720} + 1 \frac{465}{720} = 6 \frac{697}{720}$ **Ответ: 6 \frac{697}{720}**