1. Вычислите: 754 a) 9 6 15; 2 43 б) 3-98; 2 37 B) 7+8 32; 5 г) 12.(+).

а) $$\frac{7}{9} - \frac{5}{6} + \frac{4}{15}$$

Приведем дроби к общему знаменателю 90:

$$\frac{7 \cdot 10}{9 \cdot 10} - \frac{5 \cdot 15}{6 \cdot 15} + \frac{4 \cdot 6}{15 \cdot 6} = \frac{70}{90} - \frac{75}{90} + \frac{24}{90} = \frac{70 - 75 + 24}{90} = \frac{19}{90}$$

б) $$\left(2\frac{2}{3} - \frac{4}{9}\right) \cdot \frac{3}{8}$$

$$2\frac{2}{3} = \frac{2 \cdot 3 + 2}{3} = \frac{8}{3}$$

$$\frac{8}{3} - \frac{4}{9} = \frac{8 \cdot 3}{3 \cdot 3} - \frac{4}{9} = \frac{24}{9} - \frac{4}{9} = \frac{20}{9}$$

$$\frac{20}{9} \cdot \frac{3}{8} = \frac{20 \cdot 3}{9 \cdot 8} = \frac{5 \cdot 1}{3 \cdot 2} = \frac{5}{6}$$

в) $$7\frac{2}{7} + 8\frac{3}{32}$$

$$7\frac{2}{7} = \frac{7 \cdot 7 + 2}{7} = \frac{51}{7}$$

$$8\frac{3}{32} = \frac{8 \cdot 32 + 3}{32} = \frac{259}{32}$$

$$\frac{51}{7} + \frac{259}{32} = \frac{51 \cdot 32}{7 \cdot 32} + \frac{259 \cdot 7}{32 \cdot 7} = \frac{1632}{224} + \frac{1813}{224} = \frac{1632 + 1813}{224} = \frac{3445}{224} = 15\frac{65}{224}$$

г) $$\frac{5}{12} \cdot \left(\frac{1}{5} + \frac{3}{25}\right)$$

$$\frac{1}{5} + \frac{3}{25} = \frac{1 \cdot 5}{5 \cdot 5} + \frac{3}{25} = \frac{5}{25} + \frac{3}{25} = \frac{8}{25}$$

$$\frac{5}{12} \cdot \frac{8}{25} = \frac{5 \cdot 8}{12 \cdot 25} = \frac{1 \cdot 2}{3 \cdot 5} = \frac{2}{15}$$

Ответ: а) $$\frac{19}{90}$$; б) $$\frac{5}{6}$$; в) $$15\frac{65}{224}$$; г) $$\frac{2}{15}$$