Упр.546 ГДЗ Никольский Потапов 6 класс (Математика)

Задача

а) -5/9•(-18/25)-14/27•(-18/35);
б) -27/20•(-5/9)-5/24•(-22/5);
в) 21/20•(-8/21)+7/72•(-36/5);
г) -36/60•(-5/18)-(-21/56)•(-1/3).

Подробный ответ

а) $$-\frac{5}{9}\cdot\left(-\frac{18}{25}\right)-\frac{14}{27}\cdot\left(-\frac{18}{35}\right)$$

$$-\frac{5}{9}\cdot\left(-\frac{18}{25}\right)=\frac{5\cdot 18}{9\cdot 25}=\frac{2}{5}$$
$$\frac{14}{27}\cdot\left(-\frac{18}{35}\right)=-\frac{14\cdot 18}{27\cdot 35}=-\frac{4}{15}$$
$$\frac{2}{5}-\left(-\frac{4}{15}\right)=\frac{2}{5}+\frac{4}{15}=\frac{6}{15}+\frac{4}{15}=\frac{10}{15}=\frac{2}{3}$$

б) $$-\frac{27}{20}\cdot\left(-\frac{5}{9}\right)-\frac{5}{24}\cdot\left(-\frac{22}{5}\right)$$

$$-\frac{27}{20}\cdot\left(-\frac{5}{9}\right)=\frac{27\cdot 5}{20\cdot 9}=\frac{3}{4}$$
$$\frac{5}{24}\cdot\left(-\frac{22}{5}\right)=-\frac{22}{24}=-\frac{11}{12}$$
$$\frac{3}{4}-\left(-\frac{11}{12}\right)=\frac{3}{4}+\frac{11}{12}=\frac{9}{12}+\frac{11}{12}=\frac{20}{12}=\frac{5}{3}=1\frac{2}{3}$$

в) $$\frac{21}{20}\cdot\left(-\frac{8}{21}\right)+\frac{7}{72}\cdot\left(-\frac{36}{5}\right)$$

$$\frac{21}{20}\cdot\left(-\frac{8}{21}\right)=-\frac{8}{20}=-\frac{2}{5}$$
$$\frac{7}{72}\cdot\left(-\frac{36}{5}\right)=-\frac{7\cdot 36}{72\cdot 5}=-\frac{7}{10}$$
$$-\frac{2}{5}-\frac{7}{10}=-\frac{4}{10}-\frac{7}{10}=-\frac{11}{10}=-1\frac{1}{10}$$

г) $$-\frac{36}{60}\cdot\left(-\frac{5}{18}\right)-\left(-\frac{21}{56}\right)\cdot\left(-\frac{1}{3}\right)$$

$$-\frac{36}{60}\cdot\left(-\frac{5}{18}\right)=\frac{36\cdot 5}{60\cdot 18}=\frac{1}{6}$$
$$\left(-\frac{21}{56}\right)\cdot\left(-\frac{1}{3}\right)=\frac{21}{56\cdot 3}=\frac{1}{8}$$
$$\frac{1}{6}-\frac{1}{8}=\frac{4-3}{24}=\frac{1}{24}$$