Упр.849 ГДЗ Никольский Потапов 7 класс (Алгебра)

Задача

а) $$\left(3\cdot\frac{1}{3}\cdot1{,}9+19{,}5\div4\cdot\frac{1}{2}\right)\div\left(\frac{62}{75}-0{,}16\right);$$ б) $$\left(8{,}5-7\cdot\frac{3}{8}\right)\cdot5\cdot\frac{2}{3}-1{,}8\left(3\cdot\frac{1}{3}-2\cdot\frac{7}{9}\right).$$

Подробный ответ

а)

$$\left(3\frac{1}{3}\cdot 1{,}9+19{,}5:4\frac{1}{2}\right):\left(\frac{62}{75}-0{,}16\right)$$
$$=\left(\frac{10}{3}\cdot \frac{19}{10}+\frac{195}{10}:\frac{9}{2}\right):\left(\frac{62}{75}-\frac{16}{100}\right)$$
$$=\left(\frac{19}{3}+\frac{39}{2}\cdot \frac{2}{9}\right):\left(\frac{62}{75}-\frac{4}{25}\right)$$
$$=\left(\frac{19}{3}+\frac{13}{3}\right):\left(\frac{62-12}{75}\right) =\frac{32}{3}:\frac{50}{75}$$
$$=\frac{32}{3}\cdot \frac{75}{50}=16$$

б)

$$\left(8{,}5-7\frac{3}{8}\right)\cdot 5\frac{2}{3}-1{,}8\left(3\frac{1}{3}-2\frac{7}{9}\right)$$
$$=\left(8\frac{1}{2}-7\frac{3}{8}\right)\cdot 5\frac{2}{3}-1{,}8\left(3\frac{3}{9}-2\frac{7}{9}\right)$$
$$=\left(8\frac{4}{8}-7\frac{3}{8}\right)\cdot \frac{17}{3}-1{,}8\cdot \frac{3}{9}$$
$$=1\frac{1}{8}\cdot \frac{17}{3}-\frac{9}{5}\cdot \frac{1}{3} =\frac{9}{8}\cdot \frac{17}{3}-\frac{3}{5}$$
$$=\frac{51}{8}-1=5\frac{3}{8}$$