Упр.539 ГДЗ Муравин 6 класс (Математика)
- Вычислите значение выражения:
1) $$\frac{\left(6\frac{1}{2}-1\frac{3}{4}\right)\colon 2\frac{1}{2}}{\left(7-6\frac{2}{5}\right)\cdot 3\frac{1}{3}}-\frac{\frac{1}{2}+\frac{3}{4}-\frac{5}{12}}{3\frac{7}{11}+6\frac{4}{11}}$$
2) $$\frac{\left(15\colon\frac{5}{18}\colon 3\frac{3}{8}\right)\cdot\left(\frac{1}{16}+\frac{11}{36}+\frac{5}{48}+\frac{5}{18}\right)}{\left(11\frac{5}{11}-8\frac{21}{22}\right)\colon 1\frac{2}{3}}$$
3) $$\frac{\left(8\frac{1}{4}-\frac{3}{4}\right)\colon 3\frac{1}{2}}{\left(5-4\frac{2}{5}\right)\colon 10}+\frac{\left(3\frac{1}{8}-1\frac{7}{8}\right)\cdot 1\frac{3}{5}}{\left(2-1\frac{3}{8}\right)\colon 3\frac{1}{8}}$$
4) $$\frac{7\frac{1}{2}\cdot 2\frac{2}{3}-12\frac{1}{4}\colon\frac{7}{2}}{110\colon\frac{3}{5}}+\frac{3\frac{3}{8}+2\frac{3}{4}}{24\colon 2\frac{2}{5}}$$
1) Преобразуем смешанные числа в неправильные дроби и выполняем действия по порядку:
$$\frac{\left(6\frac12-1\frac34\right):2\frac12}{\left(7-6\frac25\right)\cdot 3\frac13}-\frac{\frac12+\frac34-\frac5{12}}{3\frac7{11}+6\frac4{11}} = \frac{\left(\frac{13}{2}-\frac74\right):\frac52}{\left(7-\frac{32}{5}\right)\cdot \frac{10}{3}}-\frac{\frac12+\frac34-\frac5{12}}{9\frac{11}{11}}$$
$$\frac{\frac{19}{4}:\frac52}{\frac35\cdot \frac{10}{3}}-\frac{\frac{19}{12}}{10} = \frac{\frac{19}{4}\cdot \frac25}{2}-\frac{19}{12}\cdot \frac1{10} = \frac{19}{2}\cdot \frac15\cdot \frac12-\frac1{12}$$
$$\frac{19}{20}-\frac1{12} = \frac{57-5}{60} = \frac{52}{60} = \frac{13}{15}$$
2) Сначала вычислим каждую часть:
$$15:\frac5{18}:3\frac38 = 15\cdot \frac{18}{5}\cdot \frac{8}{27} = 2\cdot 8 = 16$$
$$\frac1{16}+\frac{11}{36}+\frac5{48}+\frac5{18} = \frac3{48}+\frac{11}{36}+\frac5{48}+\frac{10}{36} = \frac8{48}+\frac{21}{36} = \frac16+\frac7{12} = \frac{9}{12} = \frac34$$
$$11\frac5{11}-8\frac{21}{22} = 11\frac{10}{22}-8\frac{21}{22} = 2\frac{11}{22} = 2\frac12$$
$$2\frac12:\,1\frac23 = \frac52:\frac53 = \frac52\cdot \frac35 = \frac32$$
Тогда
$$\frac{16\cdot \frac34}{\frac32} = 16\cdot \frac34\cdot \frac23 = 8$$
3) Вычислим по отдельности обе дроби:
$$\frac{\left(8\frac14-\frac34\right):3\frac12}{\left(5-4\frac25\right):10} = \frac{\left(7\frac12\right):3\frac12}{\frac35:10} = \frac{\frac{15}{2}:\frac72}{\frac35:\,10} = \frac{\frac{15}{2}\cdot \frac27}{\frac3{50}} = \frac{15}{7}\cdot \frac{50}{3}$$
$$= \frac{250}{7}$$
Вторая дробь:
$$\frac{\left(3\frac18-1\frac78\right)\cdot 1\frac35}{\left(2-1\frac38\right):3\frac18} = \frac{1\frac14\cdot 1\frac35}{\frac58:\frac{25}{8}} = \frac{\frac54\cdot \frac85}{\frac15} = 2\cdot 5 = 10$$
Тогда
$$\frac{250}{7}+10 = 35\frac57$$
4) Сначала найдём значения числителя и знаменателя каждой дроби:
$$7\frac12\cdot 2\frac23-12\frac14:\frac72 = \frac{15}{2}\cdot \frac83-\frac{49}{4}:\frac72 = 20-\frac72 = 16\frac12$$
$$110:\frac35 = 110\cdot \frac53 = \frac{550}{3}$$
Значит первая дробь равна
$$\frac{16\frac12}{\frac{550}{3}} = \frac{33}{2}\cdot \frac{3}{550} = \frac{9}{100}$$
Теперь вторая дробь:
$$3\frac38+2\frac34 = \frac{27}{8}+\frac{11}{4} = \frac{27}{8}+\frac{22}{8} = \frac{49}{8}$$
$$24:2\frac25 = 24:\frac{12}{5} = 24\cdot \frac5{12} = 10$$
$$\frac{\frac{49}{8}}{10} = \frac{49}{80}$$
Тогда
$$\frac{9}{100}+\frac{49}{80} = \frac{36}{400}+\frac{245}{400} = \frac{281}{400}$$
Ответ
1) $$\frac{13}{15}$$; 2) $$8$$; 3) $$35\frac57$$; 4) $$\frac{281}{400}$$.
