Ответ:
а)
\(a=-1\frac{7}{8}=-\frac{15}{8}\).
\(\frac{4}{7}\left(\frac{7}{5}\cdot\left(-\frac{15}{8}\right)-\frac{7}{2}\right)+\frac{6}{5}\left(3-2\cdot\left(-\frac{15}{8}\right)\right)\)
\(=\frac{4}{7}\left(-\frac{49}{8}\right)+\frac{6}{5}\cdot\frac{27}{4}=-\frac{7}{2}+\frac{81}{10}=\frac{23}{5}=4,6.\)
б)
Раскроем скобки:
\(\frac{5}{12}(4,8a-1,2b)-3,6\left(\frac{4}{9}a-\frac{1}{4}b\right)\)
\(=2a-\frac{1}{2}b-\frac{8}{5}a+\frac{9}{10}b=\frac{2}{5}a+\frac{2}{5}b=\frac{2}{5}(a+b).\)
При \(a+b=-2\): \(\frac{2}{5}\cdot(-2)=-\frac{4}{5}\).
Ответ: а) \(\frac{23}{5}\); б) \(-\frac{4}{5}\).
