Đáp án + Giải thích các bước giải:

Ta có: $\overrightarrow{AC} + \overrightarrow{AB'} + \overrightarrow{AD'} = (\overrightarrow{AB} + \overrightarrow{AD}) + (\overrightarrow{AB} + \overrightarrow{AA'}) + (\overrightarrow{AD} + \overrightarrow{AA'})$

$= 2\overrightarrow{AB} + 2\overrightarrow{AA'} + 2\overrightarrow{AD}$ 

$\Rightarrow \overrightarrow{c} + \overrightarrow{b} + \overrightarrow{d} = 2(\overrightarrow{AB} + \overrightarrow{AA'} + \overrightarrow{AD})$

$\Rightarrow \overrightarrow{AB} + \overrightarrow{AA'} + \overrightarrow{AD} = \dfrac{1}{2}\overrightarrow{c} + \dfrac{1}{2}\overrightarrow{b} + \dfrac{1}{2}\overrightarrow{d}$

Mà $\begin {cases} \overrightarrow{AB} + \overrightarrow{AA'} = \overrightarrow{AB'} = \overrightarrow{b} \\ \overrightarrow{AB} + \overrightarrow{AD} = \overrightarrow{AC} = \overrightarrow{c} \\ \overrightarrow{AD} + \overrightarrow{AA'} = \overrightarrow{AD'} = \overrightarrow{d} \end {cases}$

$\Rightarrow \overrightarrow{AD} + \overrightarrow{b} = \overrightarrow{AA'} + \overrightarrow{c} = \overrightarrow{AB} + \overrightarrow{d} = \dfrac{1}{2}\overrightarrow{c} + \dfrac{1}{2}\overrightarrow{b} + \dfrac{1}{2}\overrightarrow{d}$

$\Rightarrow \begin {cases} \overrightarrow{AB} = \dfrac{1}{2}\overrightarrow{c} + \dfrac{1}{2}\overrightarrow{b} - \dfrac{1}{2}\overrightarrow{d} \\ \overrightarrow{AD} = \dfrac{1}{2}\overrightarrow{c} - \dfrac{1}{2}\overrightarrow{b} + \dfrac{1}{2}\overrightarrow{d} \\ \overrightarrow{AA'} = -\dfrac{1}{2}\overrightarrow{c} + \dfrac{1}{2}\overrightarrow{b} + \dfrac{1}{2}\overrightarrow{d} \end {cases}$