Упр.160 ГДЗ Макарычев 7 класс (Алгебра)

а) (3 7/30 — 1 5/12) : 18 1/6;
б) (1 1/2 + 2 2/3) : 1 2/3;
в) (11/18 — 1 7/12) · (2 1/6 + 7/30);
г) (3 2/5 — 5) · (31/48 + 7/24).

1. $$\left(3\frac{7}{30}-1\frac{5}{12}\right):18\frac{1}{6}$$

   $$3\frac{7}{30}= \frac{97}{30}, \quad 1\frac{5}{12}= \frac{17}{12}$$

   $$\frac{97}{30}-\frac{17}{12}=\frac{194}{60}-\frac{85}{60}=\frac{109}{60}$$

   $$18\frac{1}{6}=\frac{109}{6}$$

   $$\frac{109}{60}:\frac{109}{6}=\frac{109}{60}\cdot\frac{6}{109}=\frac{1}{10}$$

2. $$\left(1\frac{1}{2}+2\frac{2}{3}\right):1\frac{2}{3}$$

   $$1\frac{1}{2}=\frac{3}{2}, \quad 2\frac{2}{3}=\frac{8}{3}$$

   $$\frac{3}{2}+\frac{8}{3}=\frac{9}{6}+\frac{16}{6}=\frac{25}{6}$$

   $$1\frac{2}{3}=\frac{5}{3}$$

   $$\frac{25}{6}:\frac{5}{3}=\frac{25}{6}\cdot\frac{3}{5}=\frac{5}{2}$$

3. $$\left(\frac{11}{18}-1\frac{7}{12}\right)\cdot\left(2\frac{1}{6}+\frac{7}{30}\right)$$

   $$1\frac{7}{12}=\frac{19}{12}$$

   $$\frac{11}{18}-\frac{19}{12}=\frac{22}{36}-\frac{57}{36}=-\frac{35}{36}$$

   $$2\frac{1}{6}=\frac{13}{6}$$

   $$\frac{13}{6}+\frac{7}{30}=\frac{65}{30}+\frac{7}{30}=\frac{72}{30}=\frac{12}{5}$$

   $$-\frac{35}{36}\cdot\frac{12}{5}=-\frac{35}{3\cdot 5}=-\frac{7}{3}$$

4. $$\left(3\frac{2}{5}-5\right)\cdot\left(\frac{31}{48}+\frac{7}{24}\right)$$

   $$3\frac{2}{5}=\frac{17}{5}$$

   $$\frac{17}{5}-5=\frac{17}{5}-\frac{25}{5}=-\frac{8}{5}$$

   $$\frac{31}{48}+\frac{7}{24}=\frac{31}{48}+\frac{14}{48}=\frac{45}{48}=\frac{15}{16}$$

   $$-\frac{8}{5}\cdot\frac{15}{16}=-\frac{3}{2}$$

Ответ

а) $$\frac{1}{10}$$; б) $$\frac{5}{2}$$; в) $$-\frac{7}{3}$$; г) $$-\frac{3}{2}$$.