53. Решите уравнение: 366. в) $$4\frac{8}{39}+\frac{7}{24} \times (1-\frac{1}{32})$$

в) $$4\frac{8}{39}+\frac{7}{24} \times (1-\frac{1}{32}) = 4\frac{8}{39}+\frac{7}{24} \times \frac{31}{32} = 4\frac{8}{39}+\frac{217}{768} = 4\frac{8 \times 256}{39 \times 256} + \frac{217 \times 39}{768 \times 39} = 4\frac{2048}{9984} + \frac{8463}{29952} = 4\frac{6144}{29952} + \frac{8463}{29952} = 4\frac{14607}{29952} = 4\frac{4869}{9984} = 4\frac{1623}{3328}$$

Ответ: $$4\frac{1623}{3328}$$