363. Раскройте скобки: 1) 6 · (2/3 a + 5/12 b); 2) 1/3 · (9/11 m - 6/7 n); 3) 12 · (3/4 x + 13/18 y - 1/24 z); 4) 1 1/7 · (7p + 21/24 q - 1 3/4).

Привет! Давай раскроем скобки, используя распределительный закон умножения: $a \cdot (b + c) = a \cdot b + a \cdot c$. **№363** 1) $6 \cdot (\frac{2}{3}a + \frac{5}{12}b) = 6 \cdot \frac{2}{3}a + 6 \cdot \frac{5}{12}b = 4a + \frac{5}{2}b = 4a + 2{,}5b$ 2) $\frac{1}{3} \cdot (\frac{9}{11}m - \frac{6}{7}n) = \frac{1}{3} \cdot \frac{9}{11}m - \frac{1}{3} \cdot \frac{6}{7}n = \frac{3}{11}m - \frac{2}{7}n$ 3) $12 \cdot (\frac{3}{4}x + \frac{13}{18}y - \frac{1}{24}z) = 12 \cdot \frac{3}{4}x + 12 \cdot \frac{13}{18}y - 12 \cdot \frac{1}{24}z = 9x + \frac{26}{3}y - \frac{1}{2}z = 9x + 8\frac{2}{3}y - 0{,}5z$ 4) $1\frac{1}{7} \cdot (7p + \frac{21}{24}q - 1\frac{3}{4}) = \frac{8}{7} \cdot 7p + \frac{8}{7} \cdot \frac{7}{8}q - \frac{8}{7} \cdot \frac{7}{4} = 8p + q - 2$ **№364** 1) $14 \cdot (\frac{1}{2}m + \frac{3}{7}n) = 14 \cdot \frac{1}{2}m + 14 \cdot \frac{3}{7}n = 7m + 6n$ 2) $\frac{1}{6} \cdot (\frac{12}{17}b - \frac{18}{23}c) = \frac{1}{6} \cdot \frac{12}{17}b - \frac{1}{6} \cdot \frac{18}{23}c = \frac{2}{17}b - \frac{3}{23}c$ 3) $8 \cdot (\frac{1}{4}p - \frac{5}{24}q + \frac{7}{12}t) = 8 \cdot \frac{1}{4}p - 8 \cdot \frac{5}{24}q + 8 \cdot \frac{7}{12}t = 2p - \frac{5}{3}q + \frac{14}{3}t = 2p - 1\frac{2}{3}q + 4\frac{2}{3}t$ 4) $1\frac{3}{4} \cdot (4a + \frac{16}{21}b - 2\frac{2}{3}) = \frac{7}{4} \cdot 4a + \frac{7}{4} \cdot \frac{16}{21}b - \frac{7}{4} \cdot \frac{8}{3} = 7a + \frac{4}{3}b - \frac{14}{3} = 7a + 1\frac{1}{3}b - 4\frac{2}{3}$