Cho DABC. Gọi M, N lần lượt là trung điểm của AB, AC.

a) Vì M là trung điểm AB nên \(\overrightarrow {CB} + \overrightarrow {CA} = 2\overrightarrow {CM} \).
b) Ta có \( - \frac{2}{3}\overrightarrow {CM} - \frac{4}{3}\overrightarrow {BN} \)\( = - \frac{2}{3}.\frac{1}{2}\left( {\overrightarrow {CB} + \overrightarrow {CA} } \right) - \frac{4}{3}.\frac{1}{2}\left( {\overrightarrow {BA} + \overrightarrow {BC} } \right)\)\( = - \frac{1}{3}\overrightarrow {CB} - \frac{1}{3}\overrightarrow {CA} - \frac{2}{3}\overrightarrow {BA} - \frac{2}{3}\overrightarrow {BC} \)
\( = \frac{1}{3}\overrightarrow {BC} + \frac{1}{3}\overrightarrow {AC} + \frac{2}{3}\overrightarrow {AB} - \frac{2}{3}\overrightarrow {BC} \)\( = - \frac{1}{3}\overrightarrow {BC} + \frac{1}{3}\overrightarrow {AC} + \frac{2}{3}\overrightarrow {AB} \)\( = \frac{1}{3}\left( {\overrightarrow {AC} - \overrightarrow {BC} } \right) + \frac{2}{3}\overrightarrow {AB} \)\( = \frac{1}{3}\left( {\overrightarrow {AC} + \overrightarrow {CB} } \right) + \frac{2}{3}\overrightarrow {AB} \)
\( = \frac{1}{3}\overrightarrow {AB} + \frac{2}{3}\overrightarrow {AB} = \overrightarrow {AB} \).
c) Ta có \(\frac{4}{3}\overrightarrow {CM} + \frac{2}{3}\overrightarrow {BN} \)\( = \frac{4}{3}.\frac{1}{2}\left( {\overrightarrow {CA} + \overrightarrow {CB} } \right) + \frac{2}{3}.\frac{1}{2}\left( {\overrightarrow {BA} + \overrightarrow {BC} } \right)\)\( = \frac{2}{3}\overrightarrow {CA} + \frac{2}{3}\overrightarrow {CB} + \frac{1}{3}\overrightarrow {BA} + \frac{1}{3}\overrightarrow {BC} \)
\( = - \frac{2}{3}\overrightarrow {AC} - \frac{1}{3}\overrightarrow {AB} + \frac{1}{3}\overrightarrow {BC} - \frac{2}{3}\overrightarrow {BC} \)\( = - \frac{2}{3}\overrightarrow {AC} - \frac{1}{3}\overrightarrow {AB} - \frac{1}{3}\overrightarrow {BC} \)\( = - \frac{2}{3}\overrightarrow {AC} - \frac{1}{3}\left( {\overrightarrow {AB} + \overrightarrow {BC} } \right)\)\( = - \frac{2}{3}\overrightarrow {AC} - \frac{1}{3}\overrightarrow {AC} = - \overrightarrow {AC} \).
d) \(\frac{1}{3}\overrightarrow {BN} - \frac{1}{3}\overrightarrow {CM} \)\( = \frac{1}{3}.\frac{1}{2}\left( {\overrightarrow {BA} + \overrightarrow {BC} } \right) - \frac{1}{3}.\frac{1}{2}\left( {\overrightarrow {CA} + \overrightarrow {CB} } \right)\)\( = \frac{1}{6}\left( {\overrightarrow {BA} + \overrightarrow {BC} - \overrightarrow {CA} - \overrightarrow {CB} } \right)\)\( = \frac{1}{2}\overrightarrow {BC} = \overrightarrow {MN} \).
Đáp án: a) Đúng; b) Đúng; c) Sai; d) Đúng.