Упр.849 ГДЗ Никольский Потапов 7 класс (Алгебра)

Задача

б) (8,5-7*3/8)*5*2/3 — 1,8(3*1/3-2*7/9).

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)

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