Упр.1008 ГДЗ Ткачёва 5 класс (Математика)

Задача

1) 1 7/9 · (5/6 — 1/3); 2) (4/9 + 1/12) · 1 1/5; 3) (2 1/2)^2 · (2/3 + 3/4);
4) (1 1/6 — 1/4) · (1 2/3)^2; 5) 2/7 + 1 1/7 · 5/6; 6) 2 1/6 — 3/8 · 1 1/5.

Подробный ответ

1) $$1\frac{7}{9}\cdot\left(\frac{5}{6}-\frac{1}{3}\right)=\frac{16}{9}\cdot\left(\frac{5}{6}-\frac{2}{6}\right)=\frac{16}{9}\cdot\frac{3}{6}=\frac{16}{9}\cdot\frac{1}{2}=\frac{8}{9}.$$

2) $$\left(\frac{4}{9}+\frac{1}{12}\right)\cdot1\frac{1}{5}=\left(\frac{16}{36}+\frac{3}{36}\right)\cdot\frac{6}{5}=\frac{19}{36}\cdot\frac{6}{5}=\frac{19}{30}.$$

3) $$\left(2\frac{1}{2}\right)^2\cdot\left(\frac{2}{3}+\frac{3}{4}\right)=\left(\frac{5}{2}\right)^2\cdot\left(\frac{8}{12}+\frac{9}{12}\right)=\frac{25}{4}\cdot\frac{17}{12}=\frac{425}{48}=8\frac{41}{48}.$$

4) $$\left(1\frac{1}{6}-\frac{1}{4}\right)\cdot\left(1\frac{2}{3}\right)^2=\left(\frac{7}{6}-\frac{1}{4}\right)\cdot\left(\frac{5}{3}\right)^2=\left(\frac{14}{12}-\frac{3}{12}\right)\cdot\frac{25}{9}=\frac{11}{12}\cdot\frac{25}{9}=\frac{275}{108}=2\frac{59}{108}.$$

5) $$\frac{2}{7}+1\frac{1}{7}\cdot\frac{5}{6}=\frac{2}{7}+\frac{8}{7}\cdot\frac{5}{6}=\frac{2}{7}+\frac{40}{42}=\frac{2}{7}+\frac{20}{21}=\frac{6}{21}+\frac{20}{21}=\frac{26}{21}=1\frac{5}{21}.$$

6) $$2\frac{1}{6}-\frac{3}{8}\cdot1\frac{1}{5}=2\frac{1}{6}-\frac{3}{8}\cdot\frac{6}{5}=2\frac{1}{6}-\frac{18}{40}=2\frac{1}{6}-\frac{9}{20}.$$

Приведём к общему знаменателю:

$$2\frac{1}{6}=\frac{13}{6}=\frac{130}{60},\qquad \frac{9}{20}=\frac{27}{60}.$$

Тогда

$$\frac{130}{60}-\frac{27}{60}=\frac{103}{60}=1\frac{43}{60}.$$