Упр.815 ГДЗ Никольский Потапов 7 класс (Алгебра)

Задача

б) (4-1*1/3*2)*1/2;
в) 6*3/5*7*1/6-2*1/6*6*3/5;
г) 3*3/4*3*3/4+3*3/4*1/2.

Подробный ответ

а)

$$ \left(4\frac{2}{8}+5\frac{1}{2}\right)\cdot 6 = \left(4\frac{2}{8}+5\frac{4}{8}\right)\cdot 6 = 9\frac{6}{8}\cdot 6 = 9\frac{3}{4}\cdot 6 = \frac{39}{4}\cdot 6 = \frac{39}{2}\cdot 3 = \frac{117}{2} = 58\frac{1}{2} $$

б)

$$ \left(4-1\frac{1}{3}\cdot 2\right)\cdot \frac{1}{2} = \left(4-\frac{4}{3}\cdot 2\right)\cdot \frac{1}{2} = \left(4-\frac{8}{3}\right)\cdot \frac{1}{2} = \left(4-2\frac{2}{3}\right)\cdot \frac{1}{2} = 1\frac{1}{3}\cdot \frac{1}{2} = \frac{4}{3}\cdot \frac{1}{2} = \frac{2}{3} $$

в)

$$ 6\frac{3}{5}\cdot 7\frac{1}{6}-2\frac{1}{6}\cdot 6\frac{3}{5} = 6\frac{3}{5}\cdot \left(7\frac{1}{6}-2\frac{1}{6}\right) = 6\frac{3}{5}\cdot 5 = \frac{33}{5}\cdot 5 = 33 $$

г)

$$ 3\frac{3}{4}\cdot 3\frac{3}{4}+3\frac{3}{4}\cdot \frac{1}{2} = 3\frac{3}{4}\cdot \left(3\frac{3}{4}+\frac{1}{2}\right) = \frac{15}{4}\cdot \left(3\frac{3}{4}+\frac{2}{4}\right) = \frac{15}{4}\cdot \frac{17}{4} = \frac{255}{16} = 15\frac{15}{16} $$