Найдите корень уравнения: а) 3/7x + 5 1/2x - 7 3/4 = 1 - 4/7x + 5 1/4

а) $\frac{3}{7}x + 5\frac{1}{2}x - 7\frac{3}{4} = 1 - \frac{4}{7}x + 5\frac{1}{4}$ $(\frac{3}{7} + \frac{4}{7})x + 5\frac{1}{2}x = 1 + 5\frac{1}{4} + 7\frac{3}{4}$ $1x + 5\frac{1}{2}x = 1 + 13$ $6\frac{1}{2}x = 14$ $x = 14 : \frac{13}{2} = 14 \cdot \frac{2}{13} = \frac{28}{13} = 2\frac{2}{13}$ **Ответ: $2\frac{2}{13}$** б) $5 - 2\frac{1}{3}z + 4\frac{4}{9}z = 7\frac{1}{2}z - 6\frac{5}{12}z + 6\frac{1}{3}$ $5 + (4\frac{4}{9} - 2\frac{3}{9})z = (7\frac{6}{12} - 6\frac{5}{12})z + 6\frac{1}{3}$ $5 + 2\frac{1}{9}z = 1\frac{1}{12}z + 6\frac{1}{3}$ $2\frac{1}{9}z - 1\frac{1}{12}z = 6\frac{1}{3} - 5$ $(\frac{19}{9} - \frac{13}{12})z = 1\frac{1}{3}$ $(\frac{76-39}{36})z = \frac{4}{3}$ $\frac{37}{36}z = \frac{4}{3}$ $z = \frac{4}{3} \cdot \frac{36}{37} = \frac{4 \cdot 12}{37} = \frac{48}{37} = 1\frac{11}{37}$ **Ответ: $1\frac{11}{37}$** в) $4 \cdot (\frac{2}{7}n + 1) + 2\frac{1}{2} = 6 - \frac{1}{3} \cdot (\frac{6}{7}n - 3)$ $\frac{8}{7}n + 4 + 2\frac{1}{2} = 6 - \frac{2}{7}n + 1$ $\frac{8}{7}n + 6\frac{1}{2} = 7 - \frac{2}{7}n$ $\frac{8}{7}n + \frac{2}{7}n = 7 - 6\frac{1}{2}$ $\frac{10}{7}n = \frac{1}{2}$ $n = \frac{1}{2} \cdot \frac{7}{10} = \frac{7}{20} = 0,35$ **Ответ: 0,35** г) $2 - (1\frac{1}{3}p + \frac{1}{7}) \cdot 21 = 4\frac{1}{4}p - 6\frac{3}{8}$ $2 - (\frac{4}{3}p \cdot 21 + \frac{1}{7} \cdot 21) = 4\frac{1}{4}p - 6\frac{3}{8}$ $2 - (28p + 3) = 4\frac{1}{4}p - 6\frac{3}{8}$ $2 - 28p - 3 = 4\frac{1}{4}p - 6\frac{3}{8}$ $-1 - 28p = 4\frac{1}{4}p - 6\frac{3}{8}$ $6\frac{3}{8} - 1 = 4\frac{1}{4}p + 28p$ $5\frac{3}{8} = 32\frac{1}{4}p$ $p = \frac{43}{8} : \frac{129}{4} = \frac{43}{8} \cdot \frac{4}{129} = \frac{1 \cdot 1}{2 \cdot 3} = \frac{1}{6}$ **Ответ: $\frac{1}{6}$**