Question

[alibaba nguyễn]

Ta có: \(\overrightarrow{B'B}=\overrightarrow{B'A}+\overrightarrow{AB}=\overrightarrow{C'D'}+\overrightarrow{DC}\)

\(\Rightarrow\overrightarrow{B'B}+\overrightarrow{CC'}+\overrightarrow{D'D}=\overrightarrow{C'D'}+\overrightarrow{DC}+\overrightarrow{CC'}+\overrightarrow{D'D}\)

\(=\left(\overrightarrow{CC'}+\overrightarrow{C'D'}\right)+\left(\overrightarrow{D'D}+\overrightarrow{DC}\right)=\overrightarrow{CD'}+\overrightarrow{D'C}=\overrightarrow{CD'}-\overrightarrow{CD'=}\overrightarrow{0}\)

Answer 1

[Dạ Vũ]

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Answer 2

[alibaba nguyễn]

À quay lại rồi