Ответ:
\(\left(\frac{6}{5}-\frac{3}{4}\right)\cdot\frac{2}{3}=\left(\frac{24}{20}-\frac{15}{20}\right)\cdot\frac{2}{3}=\frac{9}{20}\cdot\frac{2}{3}=\frac{3}{10}\).
Вынесем общий множитель \(\frac{107}{89}\):
\(\frac{107}{89}\cdot\frac{64}{70}+\frac{107}{89}\cdot\frac{25}{70}=\frac{107}{89}\left(\frac{64}{70}+\frac{25}{70}\right)=\frac{107}{89}\cdot\frac{89}{70}=\frac{107}{70}\).
\(-\frac{1}{3}\cdot\frac{6}{5}-\frac{5}{6}\cdot\frac{3}{25}=-\frac{2}{5}-\frac{1}{10}=-\frac{4}{10}-\frac{1}{10}=-\frac{1}{2}\).
\(15\left(1+\frac{1}{3}-\frac{1}{5}\right)=15\left(\frac{15}{15}+\frac{5}{15}-\frac{3}{15}\right)=15\cdot\frac{17}{15}=17\).
Ответ: \(\frac{3}{10};\ \frac{107}{70};\ -\frac{1}{2};\ 17\).
