bai1: cho tam giac ABC co $\frac{c}{b} = \frac{m_{b}}{m_{c}} \neq 1$. chung minh rang
$\cot A = \frac{1}{2}(\cot B + \cot C)$
bai2: chung minh rang tam giac ABC deu neu thoa man 1 trong cac dieu kien sau:
a/ $\left\{ \begin{array}{l} \frac{b^{3} + c^{3} - a^{3}}{b + c - a} = a^{2}\\ a = 3b\cos C \end{array} \right.$
b/ $\left\{ \begin{array}{l} \frac{b^{3} + c^{3} - a^{3}}{b + c - a} = a^{2}\\ \frac{1}{\sin B} + \frac{1}{\sin C} = \frac{4}{\sin B + \sin C}\end{array} \right.$