Заполни пропуски

Вычисли, применив распределительный закон:

В ответ запиши конечный результат вычислений.

а) \(\cfrac{1}{2}\cdot\left(-\cfrac{3}{7}\right) + \cfrac{1}{2}\cdot\left(-\cfrac{4}{7}\right) = \cfrac{1}{2}\cdot\left(-\cfrac{3}{7} + \left(-\cfrac{4}{7}\right)\right) = \cfrac{1}{2}\cdot\biggl(-1\biggr) = -\cfrac{1}{2}\) ;

б) \(\cfrac{3}{4}\cdot\cfrac{4}{17} - \cfrac{3}{4}\cdot\cfrac{21}{17} =\) [ ];

в) \(\cfrac{12}{23}\cdot\left(-\cfrac{25}{72}\right) + \cfrac{12}{23}\cdot\left(-\cfrac{11}{72}\right) =\) [ ];

г) \(\cfrac{1}{2}\cdot\left(\cfrac{3}{10} - \cfrac{2}{7}\right) - \cfrac{1}{2}\cdot\cfrac{3}{10} =\) [ ];

д) \(\cfrac{11}{20}\cdot\left(\cfrac{2}{11} - \cfrac{3}{8}\right) + \cfrac{11}{20}\cdot\cfrac{3}{8} =\) [ ];

е) \(\cfrac{3}{5}\cdot\left(-\cfrac{5}{11} + \cfrac{5}{3}\right) - \cfrac{5}{11}\cdot\left(-\cfrac{3}{5} + \cfrac{11}{5}\right) =\) [ ];

ж) \(\cfrac{11}{12}\cdot\left(\cfrac{23}{37} - \cfrac{12}{11}\right) + \cfrac{23}{37}\cdot\left(-\cfrac{11}{12} + \cfrac{37}{23}\right) =\) [ ];

з) \(\cfrac{4}{5}\cdot\left(\cfrac{6}{7} - \cfrac{5}{4}\right) + \cfrac{6}{5}\cdot\left(\cfrac{5}{6} - \cfrac{4}{7}\right) =\) [ ];

и) \(\cfrac{11}{13}\cdot\left(\cfrac{13}{22} - \cfrac{3}{7}\right) + \cfrac{3}{13}\cdot\left(-\cfrac{13}{3} + \cfrac{11}{7}\right) =\) [ ].