Упр.821 ГДЗ Никольский Потапов 7 класс (Алгебра)

Задача

б) (7*1/2*2/3 — 12*1/4:7/9):6+10*1/8;
в) (21/113 — 14/19 + 7/8-28/41) + (4/41-1/8+2/19-3/113):1/7.

Подробный ответ

а)

$$\left(5\frac{5}{7}\cdot\frac{3}{8}-5\frac{1}{4}:7\right):3+3\frac{7}{28}$$
$$=\left(\frac{40}{7}\cdot\frac{3}{8}-\frac{21}{4}:7\right):3+3\frac{7}{28}$$
$$=\left(\frac{15}{7}-\frac{3}{4}\right):3+3\frac{7}{28}$$
$$=\frac{60-21}{28}\cdot\frac{1}{3}+3\frac{7}{28}$$
$$=\frac{39}{28}\cdot\frac{1}{3}+3\frac{7}{28}=\frac{13}{28}+3\frac{7}{28}$$
$$=3\frac{20}{28}=3\frac{5}{7}$$

б)

$$\left(7\frac{1}{2}\cdot2\frac{2}{3}-12\frac{1}{4}: \frac{7}{9}\right):6+10\frac{1}{8}$$
$$=\left(\frac{15}{2}\cdot\frac{8}{3}-\frac{49}{4}\cdot\frac{9}{7}\right):6+10\frac{1}{8}$$
$$=\left(\frac{5\cdot4}{1\cdot1}-\frac{7\cdot9}{4}\right):6+10\frac{1}{8}$$
$$=\left(20-\frac{63}{4}\right):6+10\frac{1}{8}$$
$$=\left(19\frac{4}{4}-15\frac{3}{4}\right):6+10\frac{1}{8}$$
$$=4\frac{1}{4}:6+10\frac{1}{8}=\frac{17}{4}\cdot\frac{1}{6}+10\frac{1}{8}$$
$$=\frac{17}{24}+10\frac{3}{24}=10\frac{20}{24}=10\frac{5}{6}$$

в)

$$\left(\frac{21}{113}-\frac{14}{19}+\frac{7}{8}-\frac{28}{41}\right)+\left(\frac{4}{41}-\frac{1}{8}+\frac{2}{19}-\frac{3}{113}\right):\frac{1}{7}$$
$$=\frac{21}{113}-\frac{14}{19}+\frac{7}{8}-\frac{28}{41}+\left(\frac{4}{41}-\frac{1}{8}+\frac{2}{19}-\frac{3}{113}\right)\cdot 7$$
$$=\frac{21}{113}-\frac{14}{19}+\frac{7}{8}-\frac{28}{41}+\frac{28}{41}-\frac{7}{8}+\frac{14}{19}-\frac{21}{113}=0$$