Cho hình bình hành  $ABCD$  có tâm  $O$. Chứng minh rằng:
a) $\overrightarrow{CO}-\overrightarrow{OB}=\overrightarrow{BA}   $
b) $\overrightarrow{AB}-\overrightarrow{BC}=\overrightarrow{DB}   $
c) $\overrightarrow{DA}-\overrightarrow{DB}=\overrightarrow{OD}-\overrightarrow{OC}    $
d) $\overrightarrow{DA}-\overrightarrow{DB}+\overrightarrow{DC}=\overrightarrow{0}    $
a) Ta có, theo quy tắc ba điểm của phép trừ:
$\overrightarrow{BA}=\overrightarrow{OA}-\overrightarrow{OB}                                     (1)$
Mặt khác:   $\overrightarrow{OA} =\overrightarrow{CO}                (2)$
Từ  $(1),(2)$  suy ra:
$\overrightarrow{BA}=\overrightarrow{CO}-\overrightarrow{OB}   $.

b) Ta có:  $\overrightarrow{DB}=\overrightarrow{AB}-\overrightarrow{AD}       (1)$
$\overrightarrow{AD}=\overrightarrow{BC}                                (2)$
Từ $(1),(2)$  ta suy ra:  $\overrightarrow{DB}=\overrightarrow{AB}-\overrightarrow{BC}   $

c) Ta có:  $\overrightarrow{DA}-\overrightarrow{DB}=\overrightarrow{BA}           (1)$
$\overrightarrow{OD}-\overrightarrow{OC}=\overrightarrow{CD}                         (2)$
$\overrightarrow{BA}=\overrightarrow{CD}                                    (3)$
Từ  $(1),(2),(3)$  ta suy ra điều phải chứng minh.

d) $\overrightarrow{DA}-\overrightarrow{DB}+\overrightarrow{DC}=(\overrightarrow{DA}-\overrightarrow{DB}  )+\overrightarrow{DC}=\overrightarrow{BA}+\overrightarrow{DC}=\overrightarrow{BA}+\overrightarrow{AB}=\overrightarrow{BB}=\overrightarrow{0} $          $ (do  \overrightarrow{DC}=\overrightarrow{AB}  )          $.

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