Xét ΔABC có \(cosA=\dfrac{AB^2+AC^2-BC^2}{2\cdot AB\cdot AC}=\dfrac{52-BC^2}{2\cdot4\cdot6}=\dfrac{52-BC^2}{2\cdot4\cdot6}\)
=>52-BC^2=4*6=24
=>BC=2 căn 7(cm)
\(AM=\sqrt{\dfrac{4^2+6^2}{2}-\dfrac{\left(2\sqrt{7}\right)^2}{4}}=\sqrt{19}\left(cm\right)\)