Найди разность в задании 2.247, начиная с пункта а)

Конечно, помогу! а) $1 - \frac{4}{17} = \frac{17}{17} - \frac{4}{17} = \frac{17-4}{17} = \frac{13}{17}$ б) $5 - \frac{7}{13} = \frac{5 \cdot 13}{13} - \frac{7}{13} = \frac{65}{13} - \frac{7}{13} = \frac{65-7}{13} = \frac{58}{13} = 4\frac{6}{13}$ в) $6 - 5\frac{6}{11} = 6 - \frac{5 \cdot 11 + 6}{11} = 6 - \frac{55+6}{11} = 6 - \frac{61}{11} = \frac{6 \cdot 11}{11} - \frac{61}{11} = \frac{66}{11} - \frac{61}{11} = \frac{66-61}{11} = \frac{5}{11}$ г) $4\frac{5}{9} - 3 = 4\frac{5}{9} - 3\frac{0}{9} = (4-3) + \frac{5-0}{9} = 1 + \frac{5}{9} = 1\frac{5}{9}$ д) $4\frac{9}{16} - 2\frac{3}{14} = \frac{4 \cdot 16 + 9}{16} - \frac{2 \cdot 14 + 3}{14} = \frac{64+9}{16} - \frac{28+3}{14} = \frac{73}{16} - \frac{31}{14} = \frac{73 \cdot 7}{16 \cdot 7} - \frac{31 \cdot 8}{14 \cdot 8} = \frac{511}{112} - \frac{248}{112} = \frac{511-248}{112} = \frac{263}{112} = 2\frac{39}{112}$ е) $11\frac{7}{32} - 9\frac{11}{64} = \frac{11 \cdot 32 + 7}{32} - \frac{9 \cdot 64 + 11}{64} = \frac{352+7}{32} - \frac{576+11}{64} = \frac{359}{32} - \frac{587}{64} = \frac{359 \cdot 2}{32 \cdot 2} - \frac{587}{64} = \frac{718}{64} - \frac{587}{64} = \frac{718-587}{64} = \frac{131}{64} = 2\frac{3}{64}$ ж) $17\frac{2}{3} - 6\frac{7}{8} = \frac{17 \cdot 3 + 2}{3} - \frac{6 \cdot 8 + 7}{8} = \frac{51+2}{3} - \frac{48+7}{8} = \frac{53}{3} - \frac{55}{8} = \frac{53 \cdot 8}{3 \cdot 8} - \frac{55 \cdot 3}{8 \cdot 3} = \frac{424}{24} - \frac{165}{24} = \frac{424-165}{24} = \frac{259}{24} = 10\frac{19}{24}$ з) $24\frac{7}{15} - 15\frac{5}{12} = \frac{24 \cdot 15 + 7}{15} - \frac{15 \cdot 12 + 5}{12} = \frac{360+7}{15} - \frac{180+5}{12} = \frac{367}{15} - \frac{185}{12} = \frac{367 \cdot 4}{15 \cdot 4} - \frac{185 \cdot 5}{12 \cdot 5} = \frac{1468}{60} - \frac{925}{60} = \frac{1468-925}{60} = \frac{543}{60} = 9\frac{3}{60} = 9\frac{1}{20}$