vecto$BI= 2CI/3$
=> $BC= 5CI/3$
Vecto$ AI = AB+ BI = AB+ 2CI/3$
$ = AB+ \frac{2}{3} .\frac{3}{5}BC$
$= AB+\frac{2}{5} BC$
$= AB+ \frac{2}{5}(BA+AC)$
$= AB+ \frac{2}{5}BA +\frac{2}{5}AC$
$= AB - \frac{2}{5}AB -\frac{2}{5}AC$
$= \frac{3}{5}AB - \frac{2}{5}AC$
Bn ghi thêm vecto nha tại mk ko vt đc vecto