Решение:
- \( \frac{1}{2}(x - 7) + 1 = \frac{3(1-x)}{4} \)
\( \frac{x-7}{2} + 1 = \frac{3-3x}{4} \)
\( 2(x-7) + 4 = 3(1-x) \)
\( 2x - 14 + 4 = 3 - 3x \)
\( 2x - 10 = 3 - 3x \)
\( 2x + 3x = 3 + 10 \)
\( 5x = 13 \)
\( x = \frac{13}{5} \) - \( \frac{2}{5}(3-2x) = \frac{3(1+3x)}{10} - \frac{4}{5} \)
\( \frac{6-4x}{5} = \frac{3+9x}{10} - \frac{8}{10} \)
\( 2(6-4x) = 3+9x - 8 \)
\( 12 - 8x = 9x - 5 \)
\( 12 + 5 = 9x + 8x \)
\( 17 = 17x \)
\( x = 1 \) - \( \frac{1}{3}(x + 2) = \frac{2x-1}{3} \)
\( x + 2 = 2x - 1 \)
\( 2 + 1 = 2x - x \)
\( 3 = x \) - \( \frac{2-3x}{4} = \frac{3(x+1)}{8} - 1 \)
\( 2(2-3x) = 3(x+1) - 8 \)
\( 4 - 6x = 3x + 3 - 8 \)
\( 4 - 6x = 3x - 5 \)
\( 4 + 5 = 3x + 6x \)
\( 9 = 9x \)
\( x = 1 \)
Ответ: 1) \( x = \frac{13}{5} \); 2) \( x = 1 \); 3) \( x = 3 \); 4) \( x = 1 \).