\[\boxed{\mathbf{771.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[Рисунок\ по\ условию\ задачи:\]
\[Дано:\ \]
\[ABCD - параллелограмм;\ \]
\[AC \cap BD = O;\ \]
\[\overrightarrow{a} = \overrightarrow{\text{AB}};\ \]
\[\overrightarrow{b} = \overrightarrow{\text{AD}}.\]
\[Выразить:\ \]
\[\overrightarrow{\text{DC}} + \overrightarrow{\text{CB}};\ \]
\[\overrightarrow{\text{BO}} + \overrightarrow{\text{OC}};\ \]
\[\overrightarrow{\text{BO}} - \overrightarrow{\text{OC}};\ \]
\[\overrightarrow{\text{BA}} - \overrightarrow{\text{DA}}.\]
\[Решение.\]
\[1)\ \overrightarrow{\text{DC}} + \overrightarrow{\text{CB}} = \overrightarrow{\text{DB}} = \overrightarrow{\text{AB}} - \overrightarrow{\text{AD}} =\]
\[= \overrightarrow{a} - \overrightarrow{b}\ (по\ правилу\ треугольника).\]
\[2)\ \overrightarrow{\text{BO}} + \overrightarrow{\text{OC}} = \overrightarrow{\text{BC}} = \overrightarrow{\text{AD}} = \overrightarrow{b}.\]
\[3)\ \overrightarrow{\text{BO}} - \overrightarrow{\text{OC}} = \overrightarrow{\text{BO}} - \overrightarrow{\text{AO}} =\]
\[= \overrightarrow{\text{BO}} - \left( - \overrightarrow{\text{OA}} \right) = \overrightarrow{\text{BO}} + \overrightarrow{\text{OA}} =\]
\[= \overrightarrow{\text{BA}} = - \overrightarrow{\text{AB}} =\]
\[= - \overrightarrow{a}\ (по\ правилу\ треугольника).\]
\[4)\ \overrightarrow{\text{BA}} - \overrightarrow{\text{DA}} = \overrightarrow{\text{BA}} - \left( - \overrightarrow{\text{AD}} \right) =\]
\[= \overrightarrow{\text{BA}} + \overrightarrow{\text{AD}} = - \overrightarrow{\text{AB}} + \overrightarrow{\text{AD}} =\]
\[= - \overrightarrow{a} + \overrightarrow{b} = \overrightarrow{b} - \overrightarrow{a}.\]