12) $$\frac{(3\frac{4}{7}-1\frac{23}{28})-(\frac{47}{65}-\frac{29}{130})}{\frac{4}{5}-\frac{2}{5}-3\frac{3}{4}+8\frac{7}{15}-8\frac{7}{60}}$$ = $$\frac{(\frac{25}{7} - \frac{51}{28}) - (\frac{94}{130} - \frac{29}{130})}{\frac{2}{5} - \frac{15}{4} + \frac{127}{15} - \frac{487}{60}}$$ = $$\frac{(\frac{100 - 51}{28}) - (\frac{65}{130})}{\frac{24 - 225 + 508 - 487}{60}}$$ = $$\frac{\frac{49}{28} - \frac{1}{2}}{\frac{-180}{60}}$$ = $$\frac{\frac{7}{4} - \frac{1}{2}}{-3}$$ = $$\frac{\frac{7 - 2}{4}}{-3}$$ = $$\frac{\frac{5}{4}}{-3}$$ = -$$\frac{5}{12}$$
Ответ: -$$\frac{5}{12}$$