Упр.483 Часть 1 ГДЗ Виленкин Жохов 6 класс (Математика)

Задача

а) 9/56 — (7/15 — 5/12) * (3/14 + 1/2);
б) (2/3 + 7/8 — 5/6) * (1 — 5/17);
в) (2 1/2)2 * 8/15 — 5/9;
г) (2 1/3 * 1 2/7)3 * 2/9;

Подробный ответ

а) $$\frac{9}{56}-\left(\frac{7}{15}-\frac{5}{12}\right)\cdot\left(\frac{3}{14}+\frac{1}{2}\right)$$

$$\frac{7}{15}-\frac{5}{12}=\frac{28}{60}-\frac{25}{60}=\frac{3}{60}=\frac{1}{20}$$

$$\frac{3}{14}+\frac{1}{2}=\frac{3}{14}+\frac{7}{14}=\frac{10}{14}=\frac{5}{7}$$

$$\frac{9}{56}-\frac{1}{20}\cdot\frac{5}{7}=\frac{9}{56}-\frac{1}{28}=\frac{9}{56}-\frac{2}{56}=\frac{7}{56}=\frac{1}{8}$$

б) $$\left(\frac{2}{3}+\frac{7}{8}-\frac{5}{6}\right)\cdot\left(1-\frac{5}{17}\right)$$

$$\frac{2}{3}+\frac{7}{8}-\frac{5}{6}=\frac{16}{24}+\frac{21}{24}-\frac{20}{24}=\frac{17}{24}$$

$$1-\frac{5}{17}=\frac{12}{17}$$

$$\frac{17}{24}\cdot\frac{12}{17}=\frac{1}{2}$$

в) $$\left(2\frac{1}{2}\right)^2\cdot\frac{8}{15}-\frac{5}{9}$$

$$2\frac{1}{2}=\frac{5}{2}$$

$$\left(\frac{5}{2}\right)^2\cdot\frac{8}{15}-\frac{5}{9}=\frac{25}{4}\cdot\frac{8}{15}-\frac{5}{9}$$

$$\frac{25}{4}\cdot\frac{8}{15}=\frac{25\cdot 2}{15}=\frac{10}{3}$$

$$\frac{10}{3}-\frac{5}{9}=\frac{30}{9}-\frac{5}{9}=\frac{25}{9}=2\frac{7}{9}$$

г) $$\left(2\frac{1}{3}\cdot1\frac{2}{7}\right)^3\cdot\frac{2}{9}$$

$$2\frac{1}{3}=\frac{7}{3},\qquad 1\frac{2}{7}=\frac{9}{7}$$

$$\left(\frac{7}{3}\cdot\frac{9}{7}\right)^3\cdot\frac{2}{9}=3^3\cdot\frac{2}{9}$$

$$27\cdot\frac{2}{9}=6$$