Cho hình lăng tr $ABC.A'B'C'$ có:$\overrightarrow{AA'}=\overrightarrow{a}, \overrightarrow{AB}=\overrightarrow{b}, \overrightarrow{AC}=\overrightarrow{c}.$Gi $G'$ là trng tâm tam giác $A'B'C'$. Hãy biu din $\overrightarrow{AG'}$ qua các vectơ $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$.$\overrightarrow{B'C}=\overrightarrow{B'B}+\overrightarrow{BA}+\overrightarrow{AC}=-\overrightarrow{a}-\overrightarrow{b}+\overrightarrow{c}$

 
Ta có:
$3\overrightarrow{AG'}=\overrightarrow{AA'}+\overrightarrow{AB'}+\overrightarrow{AC'}$ 
Trong đó:
$\overrightarrow{AA'}=\overrightarrow{a}$ 
$\overrightarrow{AB'}=\overrightarrow{AB}+\overrightarrow{BB'}=\overrightarrow{b}+\overrightarrow{a}$
$\overrightarrow{AC'}=\overrightarrow{AC}+\overrightarrow{CC'}=\overrightarrow{c}+\overrightarrow{a} $
Từ đó suy ra: $\overrightarrow{AG'}= \overrightarrow{a}+ \frac{1}{3}(\overrightarrow{b}+\overrightarrow{c})$ 
no problem. hihi. –  khangnguyenthanh 12-10-12 08:30 PM
cảm ơn bác khang nhé –  nguyentienthanhqn 12-10-12 08:29 PM

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