Упр.671 Часть 3 ГДЗ Дорофеев Петерсон 6 класс (Математика)

Задача

а) 1/(1•2)+1/(2•3)+1/(3•4)+1/(4•5);
б) 1/(6•7)+1/(7•8)+1/(8•9)+1/(9•10);
в) 1/(1•4)+1/(4•7)+1/(7•10)+1/(10•13)+1/(13•16);
г) 1/(5•8)+1/(8•11)+1/(11•14)+1/(14•17)+1/(17•20).

Подробный ответ

а) Представим каждую дробь в виде разности соседних дробей:

$$ \frac{1}{1\cdot 2}=\frac{1}{1}-\frac{1}{2},\quad \frac{1}{2\cdot 3}=\frac{1}{2}-\frac{1}{3},\quad \frac{1}{3\cdot 4}=\frac{1}{3}-\frac{1}{4},\quad \frac{1}{4\cdot 5}=\frac{1}{4}-\frac{1}{5} $$

$$ \frac{1}{1\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{3\cdot 4}+\frac{1}{4\cdot 5} = \left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{5}\right) = 1-\frac{1}{5} = \frac{4}{5} $$

б)

$$ \frac{1}{6\cdot 7}=\frac{1}{6}-\frac{1}{7},\quad \frac{1}{7\cdot 8}=\frac{1}{7}-\frac{1}{8},\quad \frac{1}{8\cdot 9}=\frac{1}{8}-\frac{1}{9},\quad \frac{1}{9\cdot 10}=\frac{1}{9}-\frac{1}{10} $$

$$ \frac{1}{6\cdot 7}+\frac{1}{7\cdot 8}+\frac{1}{8\cdot 9}+\frac{1}{9\cdot 10} = \left(\frac{1}{6}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{9}\right)+\left(\frac{1}{9}-\frac{1}{10}\right) = \frac{1}{6}-\frac{1}{10} = \frac{5-3}{30} = \frac{1}{15} $$

в)

$$ \frac{1}{1\cdot 4}=\frac{1}{1}-\frac{1}{4},\quad \frac{1}{4\cdot 7}=\frac{1}{4}-\frac{1}{7},\quad \frac{1}{7\cdot 10}=\frac{1}{7}-\frac{1}{10},\quad \frac{1}{10\cdot 13}=\frac{1}{10}-\frac{1}{13},\quad \frac{1}{13\cdot 16}=\frac{1}{13}-\frac{1}{16} $$

$$ \frac{1}{1\cdot 4}+\frac{1}{4\cdot 7}+\frac{1}{7\cdot 10}+\frac{1}{10\cdot 13}+\frac{1}{13\cdot 16} = \left(1-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{10}\right)+\left(\frac{1}{10}-\frac{1}{13}\right)+\left(\frac{1}{13}-\frac{1}{16}\right) = 1-\frac{1}{16} = \frac{15}{16} $$

г)

$$ \frac{1}{5\cdot 8}=\frac{1}{5}-\frac{1}{8},\quad \frac{1}{8\cdot 11}=\frac{1}{8}-\frac{1}{11},\quad \frac{1}{11\cdot 14}=\frac{1}{11}-\frac{1}{14},\quad \frac{1}{14\cdot 17}=\frac{1}{14}-\frac{1}{17},\quad \frac{1}{17\cdot 20}=\frac{1}{17}-\frac{1}{20} $$

$$ \frac{1}{5\cdot 8}+\frac{1}{8\cdot 11}+\frac{1}{11\cdot 14}+\frac{1}{14\cdot 17}+\frac{1}{17\cdot 20} = \left(\frac{1}{5}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{11}\right)+\left(\frac{1}{11}-\frac{1}{14}\right)+\left(\frac{1}{14}-\frac{1}{17}\right)+\left(\frac{1}{17}-\frac{1}{20}\right) = \frac{1}{5}-\frac{1}{20} = \frac{3}{20} $$