Задание 159 Вариант 3 Самостоятельная работа ГДЗ Дидактические материалы Чесноков 6 класс (Математика)

а) $$\frac{5{,}7}{22{,}8}=\frac{57}{228}=\frac{1}{4};$$

б) $$\frac{1\frac{1}{4}}{1\frac{1}{2}}=\frac{5}{4}:\frac{3}{2}=\frac{5}{4}\cdot\frac{2}{3}=\frac{5\cdot2}{4\cdot3}=\frac{5}{6};$$

в) $$\frac{4{,}6\cdot2{,}1\cdot0{,}5}{3{,}5\cdot2{,}3\cdot0{,}3}=\frac{46\cdot21\cdot5}{35\cdot23\cdot3}=\frac{2\cdot7\cdot1}{7\cdot1\cdot1}=2;$$

г) $$\frac{6{,}8\cdot0{,}04\cdot1{,}65}{3{,}3\cdot5{,}1\cdot0{,}16}=\frac{68\cdot4\cdot165}{33\cdot51\cdot16\cdot10}=\frac{4\cdot1\cdot5}{1\cdot3\cdot4\cdot10}=\frac{1}{6};$$

д) $$\frac{0{,}804:1{,}2+0{,}33}{9{,}5:2{,}4-3\frac{11}{12}}=\frac{0{,}67+0{,}33}{\frac{95}{24}-3\frac{11}{12}}=\frac{1}{3\frac{23}{24}-3\frac{22}{24}}=\frac{1}{\frac{1}{24}}=24;$$

е) $$\frac{\left(3\frac{1}{2}+1\frac{4}{5}\right)\cdot\frac{3}{5}-1{,}18}{8\frac{1}{8}\cdot\frac{2}{13}+1\frac{1}{3}:1\frac{7}{9}}=\frac{\left(3\frac{5}{10}+1\frac{8}{10}\right)\cdot\frac{3}{5}-1{,}18}{\frac{65}{8}\cdot\frac{2}{13}+\frac{4}{3}:\frac{16}{9}}=\frac{4\frac{13}{10}\cdot\frac{3}{5}-1{,}18}{\frac{65}{8}\cdot\frac{2}{13}+\frac{4}{3}\cdot\frac{9}{16}}=\frac{\frac{53}{10}\cdot\frac{3}{5}-1{,}18}{\frac{5}{4}\cdot1+\frac{1}{3}\cdot4}=\frac{\frac{159}{50}-1{,}18}{\frac{5}{4}+\frac{3}{4}}=\frac{3{,}18-1{,}18}{\frac{8}{4}}=\frac{2}{2}=1;$$

Ответ: а) $$\frac{1}{4}$$; б) $$\frac{5}{6}$$; в) $$2$$; г) $$\frac{1}{6}$$; д) $$24$$; е) $$1$$.