Задание 105 Вариант 4 Самостоятельная работа ГДЗ Дидактические материалы Чесноков 6 класс (Математика)

а)

$$\left(1\frac{17}{25}\cdot 2\frac{1}{7}-2\frac{4}{7}\cdot 1\frac{2}{5}\right)\cdot 2\frac{7}{9}=0$$
$$1\frac{17}{25}\cdot 2\frac{1}{7}=\frac{42}{25}\cdot\frac{15}{7}=\frac{42\cdot 15}{25\cdot 7}=\frac{6\cdot 3}{5\cdot 1}=\frac{18}{5}=3\frac{3}{5}$$
$$2\frac{4}{7}\cdot 1\frac{2}{5}=\frac{18}{7}\cdot\frac{7}{5}=\frac{18}{5}=3\frac{3}{5}$$
$$3\frac{3}{5}-3\frac{3}{5}=0,\qquad 0\cdot 2\frac{7}{9}=0$$

б)

$$\frac{1}{13}\cdot\left(2\frac{3}{8}-1\frac{5}{6}\right)\cdot 2\frac{2}{5}+\frac{9}{10}=1$$
$$2\frac{3}{8}-1\frac{5}{6}=\frac{2\cdot 24+9}{24}-\frac{1\cdot 24+20}{24}=\frac{57}{24}-\frac{44}{24}=\frac{13}{24}$$
$$\frac{1}{13}\cdot\frac{13}{24}=\frac{1}{24}$$
$$\frac{1}{24}\cdot 2\frac{2}{5}=\frac{1}{24}\cdot\frac{12}{5}=\frac{1}{2\cdot 5}=\frac{1}{10}$$
$$\frac{1}{10}+\frac{9}{10}=1$$

в)

$$\left(4\frac{3}{4}-2\frac{2}{5}\cdot 2\frac{1}{2}\right)\cdot 1\frac{3}{5}=2$$
$$2\frac{2}{5}\cdot 2\frac{1}{2}=\frac{12}{5}\cdot\frac{5}{2}=\frac{12}{2}=6$$
$$4\frac{3}{4}-6=\frac{19}{4}-\frac{24}{4}=-\frac{5}{4}$$
$$-\frac{5}{4}\cdot 1\frac{3}{5}=-\frac{5}{4}\cdot\frac{8}{5}=-2$$

г)

$$1\frac{1}{2}\cdot\frac{2}{9}-\left(6-1\frac{3}{5}\right)^2=\frac{13}{75}$$
$$6-1\frac{3}{5}=6-\frac{8}{5}=\frac{30}{5}-\frac{8}{5}=\frac{22}{5}$$
$$\left(\frac{22}{5}\right)^2=\frac{484}{25}$$
$$1\frac{1}{2}\cdot\frac{2}{9}=\frac{3}{2}\cdot\frac{2}{9}=\frac{1}{3}$$
$$\frac{1}{3}-\frac{4}{25}=\frac{25}{75}-\frac{12}{75}=\frac{13}{75}$$

д)

$$\left(4\frac{1}{15}-1\frac{9}{10}\right)\cdot 6\frac{6}{7}+3^2=3\frac{1}{7}$$
$$4\frac{1}{15}-1\frac{9}{10}=4\frac{2}{30}-3\frac{27}{30}=3\frac{32}{30}-3\frac{27}{30}=\frac{5}{30}=\frac{1}{6}$$
$$\frac{1}{6}\cdot 6\frac{6}{7}=\frac{1}{6}\cdot\frac{48}{7}=\frac{8}{7}=1\frac{1}{7}$$
$$1\frac{1}{7}+2=3\frac{1}{7}$$

е)

$$\frac{1}{3}\cdot\left(4-\left(1\frac{1}{2}\right)^3\right)\cdot\left(3-2\frac{7}{15}\right)=\frac{1}{9}$$
$$\left(1\frac{1}{2}\right)^3=\left(\frac{3}{2}\right)^3=\frac{27}{8}=3\frac{3}{8}$$
$$4-3\frac{3}{8}=4\frac{8}{8}-3\frac{3}{8}=\frac{5}{8}$$
$$3-2\frac{7}{15}=2\frac{15}{15}-2\frac{7}{15}=\frac{8}{15}$$
$$\frac{1}{3}\cdot\frac{5}{8}\cdot\frac{8}{15}=\frac{1\cdot 5\cdot 8}{3\cdot 8\cdot 15}=\frac{1}{9}$$