Упр.1135 ГДЗ Никольский Потапов 6 класс (Математика)

1. Найдите значение числового выражения: а) $$3\frac{3}{7}\cdot3\frac{1}{2}\colon\left(1\frac{1}{11}-\frac{27}{55}\right)$$; б) $$\left(2\frac{1}{2}\colon10+10\colon2\frac{1}{2}-2\frac{1}{6}\right)\cdot\frac{36}{125}$$; в) $$3\frac{1}{8}\colon\left(\left(4\frac{5}{12}-3\frac{13}{24}\right)\cdot\frac{4}{7}+\left(3\frac{1}{18}-2\frac{7}{12}\right)\cdot1\frac{10}{17}\right)$$.

а) Преобразуем смешанные числа в неправильные дроби:

$$3\frac{3}{7}\cdot 3\frac{1}{2}:\left(1\frac{1}{11}-\frac{27}{55}\right) = \frac{24}{7}\cdot \frac{7}{2}:\left(\frac{12}{11}-\frac{27}{55}\right)$$

$$\frac{12}{11}-\frac{27}{55}=\frac{60}{55}-\frac{27}{55}=\frac{33}{55}=\frac{3}{5}$$

$$\frac{24}{7}\cdot \frac{7}{2}:\frac{3}{5} =12:\frac{3}{5} =12\cdot \frac{5}{3} =20$$

б)

$$\left(2\frac{1}{2}:10+10:2\frac{1}{2}-2\frac{1}{6}\right)\cdot \frac{36}{125}$$

$$\left(\frac{5}{2}:10+10:\frac{5}{2}-2\frac{1}{6}\right)\cdot \frac{36}{125} = \left(\frac{5}{2}\cdot \frac{1}{10}+10\cdot \frac{2}{5}-2\frac{1}{6}\right)\cdot \frac{36}{125}$$

$$\left(\frac{1}{4}+4-2\frac{1}{6}\right)\cdot \frac{36}{125} = \left(\frac{1}{4}+\frac{24}{6}-\frac{13}{6}\right)\cdot \frac{36}{125} = \left(\frac{1}{4}+\frac{11}{6}\right)\cdot \frac{36}{125}$$

$$\frac{1}{4}+\frac{11}{6} = \frac{3}{12}+\frac{22}{12} = \frac{25}{12}$$

$$\frac{25}{12}\cdot \frac{36}{125} = \frac{3}{1}\cdot \frac{1}{5} = \frac{3}{5}$$

в)

$$3\frac{1}{8}:\left(\left(4\frac{5}{12}-3\frac{13}{24}\right)\cdot \frac{4}{7} +\left(3\frac{1}{18}-2\frac{7}{12}\right)\cdot 1\frac{10}{17}\right)$$

$$3\frac{1}{8}=\frac{25}{8},\quad 4\frac{5}{12}=\frac{53}{12},\quad 3\frac{13}{24}=\frac{85}{24},\quad 3\frac{1}{18}=\frac{55}{18},\quad 2\frac{7}{12}=\frac{31}{12},\quad 1\frac{10}{17}=\frac{27}{17}$$

$$\left(\frac{53}{12}-\frac{85}{24}\right)\cdot \frac{4}{7} = \left(\frac{106}{24}-\frac{85}{24}\right)\cdot \frac{4}{7} = \frac{21}{24}\cdot \frac{4}{7} = \frac{3}{4}\cdot \frac{4}{7} = \frac{3}{7}$$

$$\left(\frac{55}{18}-\frac{31}{12}\right)\cdot \frac{27}{17} = \left(\frac{110}{36}-\frac{93}{36}\right)\cdot \frac{27}{17} = \frac{17}{36}\cdot \frac{27}{17} = \frac{3}{4}$$

Тогда

$$\frac{25}{8}:\left(\frac{3}{7}+\frac{3}{4}\right) = \frac{25}{8}:\left(\frac{12}{28}+\frac{21}{28}\right) = \frac{25}{8}:\frac{33}{28} = \frac{25}{8}\cdot \frac{28}{33} = \frac{25\cdot 7}{2\cdot 33} = \frac{175}{66}$$

Ответ

а) 20; б) $$\frac{3}{5}$$; в) $$\frac{175}{66}$$.