Упр.822 ГДЗ Никольский Потапов 7 класс (Алгебра)

б) (4/15+7/12)/(23/40-1);
в) (36*2/3:15 + 8*2/3*7)/(12*1/3 + 8*6/7:2*4/7);
г) (2*3/8:3/4 — 24*7/9)/(7*2/3+2:24).

1. а)

   $$ \frac{\frac{5}{14}-\frac{8}{21}}{\frac{16}{21}-1} = \frac{\frac{15-16}{42}}{\frac{16-21}{21}} = \frac{-\frac{1}{42}}{-\frac{5}{21}} = \frac{1}{42}\cdot\frac{21}{5} = \frac{1}{10} $$

2. б)

   $$ \frac{\frac{4}{15}+\frac{7}{12}}{\frac{23}{40}-1} = \frac{\frac{16+35}{60}}{\frac{23-40}{40}} = \frac{\frac{51}{60}}{-\frac{17}{40}} = \frac{17}{20}\cdot\left(-\frac{40}{17}\right) = -2 $$

3. в)

   $$ \frac{36\frac{2}{3}:15+8\frac{2}{3}\cdot 7}{12\frac{1}{3}+8\frac{6}{7}:2\frac{4}{7}} = \frac{\frac{110}{3}\cdot\frac{1}{15}+\frac{26}{3}\cdot 7}{12\frac{1}{3}+\frac{62}{7}:\frac{18}{7}} = \frac{\frac{22}{3}+\frac{182}{3}}{12\frac{1}{3}+\frac{62}{7}\cdot\frac{7}{18}} = \frac{\frac{204}{3}}{12\frac{1}{3}+\frac{31}{9}} = \frac{68}{12\frac{1}{3}+\frac{31}{9}} $$

   $$ 12\frac{1}{3}=\frac{37}{3}=\frac{111}{9}, \qquad \frac{111}{9}+\frac{31}{9}=\frac{142}{9} $$

   $$ \frac{68}{\frac{142}{9}}=68\cdot\frac{9}{142}=4 $$

4. г)

   $$ \frac{2\frac{3}{8}: \frac{3}{4}-24\cdot\frac{7}{9}}{7\frac{2}{3}+2:24} = \frac{\frac{19}{8}\cdot\frac{4}{3}-8\cdot\frac{7}{3}}{7\frac{2}{3}+\frac{1}{12}} = \frac{\frac{19}{6}-\frac{56}{3}}{7\frac{8}{12}+\frac{1}{12}} = \frac{\frac{19-112}{6}}{7\frac{9}{12}} = \frac{-\frac{93}{6}}{7\frac{3}{4}} $$

   $$ 7\frac{3}{4}=\frac{31}{4}, \qquad -\frac{93}{6}:\frac{31}{4} = -\frac{93}{6}\cdot\frac{4}{31} = -\frac{3}{3}\cdot\frac{2}{1} = -2 $$

Ответ

а) $$\frac{1}{10}$$; б) $$-2$$; в) $$4$$; г) $$-2$$.