á) 1-1,5δ+3\(\frac{2}{5}\)δ=1\(\frac{1}{3}\)δ-2\(\frac{7}{15}\)δ+2\(\frac{1}{2}\)

Решение:

1. \( 1 - \frac{3}{2} \delta + \frac{17}{5} \delta = \frac{4}{3} \delta - \frac{37}{15} \delta + \frac{5}{2} \)
2. \( 1 + \delta (\frac{17}{5} - \frac{3}{2}) = \delta (\frac{4}{3} - \frac{37}{15}) + \frac{5}{2} \)
3. \( 1 + \delta (\frac{34-15}{10}) = \delta (\frac{20-37}{15}) + \frac{5}{2} \)
4. \( 1 + \frac{19}{10} \delta = -\frac{17}{15} \delta + \frac{5}{2} \)
5. \( \frac{19}{10} \delta + \frac{17}{15} \delta = \frac{5}{2} - 1 \)
6. \( \delta (\frac{57+34}{30}) = \frac{3}{2} \)
7. \( \frac{91}{30} \delta = \frac{3}{2} \)
8. \( \delta = \frac{3}{2} \cdot \frac{30}{91} \)
9. \( \delta = \frac{90}{182} = \frac{45}{91} \)

Ответ: \( \delta = \frac{45}{91} \).