Разложить рациональное число в конечную цепную дробь

а) 19/37 ; б) 49/30 ; в) 81/71 .

А) 19/37:

$$
\frac{19}{37} = \frac{1}{1+\frac{18}{37}} = \frac{1}{1+\frac{1}{\frac{37}{18}}} = \frac{1}{1+\frac{1}{2+\frac{1}{18}}} = [0;1,2,18]
$$

б) 49/30:

$$
\frac{49}{30} = 1 + \frac{19}{30} = 1 + \frac{1}{\frac{30}{19}} = 1 + \frac{1}{1+\frac{11}{19}} = 1 + \frac{1}{1+\frac{1}{\frac{19}{11}}} = 1 + \frac{1}{1+\frac{1}{1+\frac{8}{11}}} = [1;1,1,8]
$$

в) 81/71:

$$
\frac{81}{71} = 1 + \frac{10}{71} = 1 + \frac{1}{\frac{71}{10}} = 1 + \frac{1}{7+\frac{1}{10}} = 1 + \frac{1}{7+\frac{1}{1+\frac{9}{10}}} = 1 + \frac{1}{7+\frac{1}{1+\frac{1}{\frac{10}{9}}}} = 1 + \frac{1}{7+\frac{1}{1+\frac{1}{1+\frac{1}{9}}}} = [1;7,1,1,9]
$$