\(\overrightarrow{AB}\cdot\overrightarrow{AC}=2\cdot3\cdot cos120=-3\)
Xét ΔBAC có \(cosA=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}\)
=>\(\dfrac{2^2+3^2-BC^2}{2\cdot2\cdot3}=\dfrac{-1}{2}\)
\(\Leftrightarrow13-BC^2=-\dfrac{12}{2}=-6\)
=>BC^2=19
=>\(BC=\sqrt{19}\left(cm\right)\)
\(AM=\sqrt{\dfrac{2\left(2^2+3^2\right)}{2}-\dfrac{19}{4}}=\dfrac{\sqrt{33}}{2}\left(cm\right)\)