Ta có: \frac{7k + 7k + 2}{7} = (14.k + 2) / 7
Ta có: a = 7kb = 7k + 2c = 7k + 3Câu a\frac{7k+7k+2}{7} = \frac{14k+2}{7}. vậy: dư 2Câu b\frac{7k+2+7k+3}{7k+3} = \frac{14k+5}{7}. vậy: dư 5\frac{7k + 7k + 2}{7} = (14.k + 2) / 7
Ta có: \frac{7k + 7k + 2}{7} = (14.k + 2) / 7