\[\boxed{\mathbf{915.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[\mathbf{Дано:}\]
\[ABCD - параллелограмм;\]
\[AM\ :MC = 4\ :1;\]
\[M \in AC.\]
\[\mathbf{Найти:}\]
\[\overrightarrow{\text{AM}}\ по\ \frac{\overrightarrow{a} = \overrightarrow{\text{AB}}}{\overrightarrow{b} = \overrightarrow{\text{AD}}}.\]
\[\mathbf{Решение.}\]
\[\overrightarrow{\text{AM}} \uparrow \uparrow \overrightarrow{\text{AC}};\left| \overrightarrow{\text{AM}} \right| = \frac{4}{5}\left| \overrightarrow{\text{AC}} \right|:\]
\[\overrightarrow{\text{AC}} = \overrightarrow{\text{AB}} + \overrightarrow{\text{AD}}\ \]
\[(по\ правилу\ параллелограмма).\]
\[Следовательно:\ \]
\[\overrightarrow{\text{AM}} = \frac{4}{5}\left( \overrightarrow{\text{AB}} + \overrightarrow{\text{AD}} \right) = \frac{4}{5}\left( \overrightarrow{a} + \overrightarrow{b} \right).\]