Cho ak va bm la hai dg trung tuyen cua tam giac abc hay phan tich cac vecto ab bc ca theo hai vecto u bang ak v bang bm

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1. $$\eqalign{
   & \overrightarrow {AB} = \overrightarrow {AK} + \overrightarrow {KM} + \overrightarrow {MB} \cr
   & = \overrightarrow {AK} – {1 \over 2}\overrightarrow {AB} – \overrightarrow {BM} \,\,\left( {KM\,\,la\,\,duong\,\,TB\,\,cua\,\,\Delta ABC} \right) \cr
   & \Leftrightarrow {3 \over 2}\overrightarrow {AB} = \overrightarrow {AK} – \overrightarrow {BM} \cr
   & \Leftrightarrow \overrightarrow {AB} = {2 \over 3}\left( {\overrightarrow {AK} – \overrightarrow {BM} } \right) \cr
   & \Leftrightarrow \overrightarrow {AB} = {2 \over 3}\left( {\overrightarrow u – \overrightarrow v } \right) \cr
   & \cr
   & \overrightarrow {AC} = 2\overrightarrow {AK} – \overrightarrow {AB} = 2\overrightarrow u – {2 \over 3}\left( {\overrightarrow u – \overrightarrow v } \right) \cr
   & \,\,\,\,\,\,\,\,\, = 2\overrightarrow u – {2 \over 3}\overrightarrow u + {2 \over 3}\overrightarrow v = {4 \over 3}\overrightarrow u + {2 \over 3}\overrightarrow v \cr
   & \cr
   & \overrightarrow {BC} = \overrightarrow {AC} – \overrightarrow {AB} = {4 \over 3}\overrightarrow u + {2 \over 3}\overrightarrow v – {2 \over 3}\left( {\overrightarrow u – \overrightarrow v } \right) \cr
   & \,\,\,\,\,\,\,\,\, = {4 \over 3}\overrightarrow u + {2 \over 3}\overrightarrow v – {2 \over 3}\overrightarrow u + {2 \over 3}\overrightarrow v = 3\overrightarrow v \cr} $$

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