a)
$\begin{array}{l} \overrightarrow {AH} = \overrightarrow {AB} + \overrightarrow {BH} = \overrightarrow {AB} + 2\overrightarrow {BG} = \overrightarrow {AB} + \dfrac{4}{3}\overrightarrow {BN} \\ = \overrightarrow {AB} + \dfrac{4}{3}.\dfrac{1}{2}\left( {\overrightarrow {BA} + \overrightarrow {BC} } \right) = \overrightarrow {AB} + \dfrac{2}{3}\left( { - \overrightarrow {AB} + \overrightarrow {AC} - \overrightarrow {AB} } \right)\\ = \overrightarrow {AB} - \dfrac{2}{3}\overrightarrow {AB} + \dfrac{2}{3}\overrightarrow {AC} - \dfrac{2}{3}\overrightarrow {AB} = - \dfrac{1}{3}\overrightarrow {AB} + \dfrac{2}{3}\overrightarrow {AC} \end{array}$
b)
$\overrightarrow {CH} = \overrightarrow {AH} - \overrightarrow {AC} = - \dfrac{1}{3}\overrightarrow {AB} + \dfrac{2}{3}\overrightarrow {AC} - \overrightarrow {AC} = - \dfrac{1}{3}\overrightarrow {AB} - \dfrac{1}{3}\overrightarrow {AC} $
c)
Ta có:
$\begin{array}{l} \overrightarrow {MH} = \overrightarrow {MB} + \overrightarrow {BH} = \dfrac{1}{2}\overrightarrow {CB} + 2\overrightarrow {BG} = \dfrac{1}{2}\left( {\overrightarrow {AB} - \overrightarrow {AC} } \right) + \dfrac{4}{3}\overrightarrow {BN} \\ = \dfrac{1}{2}\overrightarrow {AB} - \dfrac{1}{2}\overrightarrow {AC} + \dfrac{4}{3}.\dfrac{1}{2}\left( {\overrightarrow {BA} + \overrightarrow {BC} } \right)\\ = \dfrac{1}{2}\overrightarrow {AB} - \dfrac{1}{2}\overrightarrow {AC} + \dfrac{2}{3}\left( { - \overrightarrow {AB} + \overrightarrow {AC} - \overrightarrow {AB} } \right)\\ = \dfrac{1}{2}\overrightarrow {AB} - \dfrac{1}{2}\overrightarrow {AC} - \dfrac{2}{3}\overrightarrow {AB} + \dfrac{2}{3}\overrightarrow {AC} - \dfrac{2}{3}\overrightarrow {AB} \\ = - \dfrac{5}{6}\overrightarrow {AB} + \dfrac{1}{6}\overrightarrow {AC} = \dfrac{1}{6}\left( { - 5\overrightarrow {AB} + \overrightarrow {AC} } \right) \end{array}$
