Упр.895 ГДЗ Никольский Потапов 6 класс (Математика)

1. Вычислите:
   
   а) $$\frac{(-0{,}125\cdot5)}{\left(1\frac{28}{63}-\frac{17}{21}\right)\cdot7\frac{7}{8}}$$;
   
   б) $$\frac{\left(\frac{21}{40}-\frac{19}{24}\right)\cdot0{,}7+0{,}04}{0{,}675\cdot3{,}4-2{,}02}$$;
   
   в) $$\frac{\left(13{,}25-2\frac{5}{27}-10\frac{5}{6}\right)\cdot230{,}04+46{,}75}{\left(1\frac{3}{7}+3\frac{1}{3}\right)\div\left(12\frac{1}{3}-14\frac{2}{7}\right)}$$.

а) Преобразуем смешанные числа и выполняем деление в знаменателе:

$$\frac{-0{,}125\cdot 5}{\left(1\frac{28}{63}-\frac{17}{21}\right)\cdot 7\frac{7}{8}} = \frac{-0{,}125\cdot 5}{\left(\frac{91}{63}-\frac{51}{63}\right)\cdot \frac{63}{8}} = \frac{-0{,}125\cdot 5}{\frac{40}{63}\cdot \frac{63}{8}} = \frac{-0{,}125\cdot 5}{5} = -0{,}125$$

б) Сначала вычислим числитель и знаменатель:

$$\frac{\left(\frac{21}{40}-\frac{19}{24}\right)\cdot 0{,}7+0{,}04}{0{,}675\cdot 3{,}4-2{,}02} = \frac{\left(\frac{63}{120}-\frac{95}{120}\right)\cdot \frac{7}{10}+\frac{4}{100}}{2{,}295-2{,}02} = \frac{-\frac{32}{120}\cdot \frac{7}{10}+\frac{4}{100}}{0{,}275}$$

$$= \frac{-\frac{4}{15}\cdot \frac{7}{10}+\frac{4}{100}}{0{,}275} = \frac{-\frac{2}{15}\cdot \frac{7}{5}+\frac{1}{25}}{0{,}275} = \frac{-\frac{14}{75}+\frac{3}{75}}{0{,}275} = \frac{-\frac{11}{75}}{0{,}275}$$

$$0{,}275=\frac{11}{40}, \quad \frac{-\frac{11}{75}}{\frac{11}{40}} = -\frac{11}{75}\cdot \frac{40}{11} = -\frac{40}{75} = -\frac{8}{15}$$

в) Упростим выражение по частям:

$$\frac{\left(13{,}25-2\frac{5}{27}-10\frac{5}{6}\right)\cdot 230{,}04+46{,}75}{\left(1\frac{3}{7}+3\frac{1}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}$$

$$13{,}25=13\frac{1}{4}, \quad 2\frac{5}{27}=2\frac{10}{54}, \quad 10\frac{5}{6}=10\frac{45}{54}$$

$$13\frac{1}{4}-2\frac{10}{54}-10\frac{45}{54} = 13\frac{1}{4}-12\frac{55}{54} = \frac{25}{108}$$

$$230{,}04=\frac{23004}{100}, \quad 46{,}75=\frac{4675}{100}$$

$$\left(\frac{25}{108}\right)\cdot \frac{23004}{100}+46{,}75 = 53{,}25+46{,}75 = 100$$

Теперь знаменатель:

$$1\frac{3}{7}+3\frac{1}{3} = 1\frac{9}{21}+3\frac{7}{21} = 4\frac{16}{21}$$

$$12\frac{1}{3}-14\frac{2}{7} = 12\frac{7}{21}-14\frac{6}{21} = -1\frac{20}{21} = -\frac{41}{21}$$

$$\left(4\frac{16}{21}\right):\left(-\frac{41}{21}\right) = \frac{100}{21}:\left(-\frac{41}{21}\right) = -\frac{100}{41}$$

Тогда

$$100:\left(-\frac{100}{41}\right) = 100\cdot \left(-\frac{41}{100}\right) = -41$$

Ответ

а) $$-0{,}125$$; б) $$-\frac{8}{15}$$; в) $$-41$$.