\[\boxed{\mathbf{617.}ОК\ ГДЗ\ –\ домашка\ на\ 5}\]
\[Дано:\]
\[Решение.\]
\[Воспользуемся\ сочетательным\ \]
\[и\ переместительным\ законами;\]
\[и\ тем,\ что\ противоположные\ \]
\[ребра\ параллелепипеда\ равны\ \]
\[и\ параллельны.\]
\[\textbf{а)}\ \overrightarrow{\text{AB}} + \overrightarrow{B_{1}C_{1}} + \overrightarrow{DD_{1}} + \overrightarrow{\text{CD}} =\]
\[= \left( \overrightarrow{\text{AB}} + \overrightarrow{B_{1}C_{1}} \right) + \left( \overrightarrow{DD_{1}} + \overrightarrow{\text{CD}} \right) =\]
\[= \left( \overrightarrow{\text{AB}} + \overrightarrow{\text{BC}} \right) + \left( \overrightarrow{\text{CD}} + \overrightarrow{DD_{1}} \right) =\]
\[= \overrightarrow{\text{AC}} + \overrightarrow{CD_{1}} = \overrightarrow{AD_{1}}.\]
\[\textbf{в)}\ По\ правилу\ \]
\[параллелограмма:\]
\[\overrightarrow{\text{BA}} + \overrightarrow{\text{AC}} + \overrightarrow{\text{CB}} + \overrightarrow{\text{DC}} + \overrightarrow{\text{DA}} =\]
\[= \left( \overrightarrow{\text{DA}} + \overrightarrow{\text{AC}} \right) + \left( \overrightarrow{\text{DC}} + \overrightarrow{\text{CB}} \right) + \overrightarrow{\text{BA}} =\]
\[= \overrightarrow{\text{DC}} + \overrightarrow{\text{DB}} + \overrightarrow{\text{BA}} = \overrightarrow{\text{DC}} + \overrightarrow{\text{DA}} =\]
\[= \overrightarrow{\text{DB}}.\]