Решение задания 2.246

a) $$38\frac{1}{8} + 4\frac{5}{24} = 38\frac{3}{24} + 4\frac{5}{24} = (38+4) + (\frac{3}{24} + \frac{5}{24}) = 42 + \frac{8}{24} = 42 + \frac{1}{3} = 42\frac{1}{3}$$

б) $$4\frac{19}{75} + 6\frac{17}{45} = 4\frac{19*3}{75*3} + 6\frac{17*5}{45*5} = 4\frac{57}{225} + 6\frac{85}{225} = (4+6) + (\frac{57}{225} + \frac{85}{225}) = 10 + \frac{142}{225} = 10\frac{142}{225}$$

в) $$47\frac{3}{14} + 1\frac{8}{21} = 47\frac{3*3}{14*3} + 1\frac{8*2}{21*2} = 47\frac{9}{42} + 1\frac{16}{42} = (47+1) + (\frac{9}{42} + \frac{16}{42}) = 48 + \frac{25}{42} = 48\frac{25}{42}$$

г) $$54\frac{3}{4} + 18\frac{5}{6} = 54\frac{3*3}{4*3} + 18\frac{5*2}{6*2} = 54\frac{9}{12} + 18\frac{10}{12} = (54+18) + (\frac{9}{12} + \frac{10}{12}) = 72 + \frac{19}{12} = 72 + 1\frac{7}{12} = 73\frac{7}{12}$$