Mình đang cần gấp ạ chỉ cần giải câu a và b thui ạ tại mình bí hai câu đó ạ

$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\overrightarrow{CM} =\frac{1}{2}(\overrightarrow{CA} +\overrightarrow{CB})\\
\overrightarrow{BN} =\frac{1}{2}(\overrightarrow{BA} +\overrightarrow{BC})
\end{array}$
a) Ta có:
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
-\frac{2}{3}\overrightarrow{CM} -\frac{4}{3}\overrightarrow{BN} =\frac{1}{3}( -2\overrightarrow{CM} -4\overrightarrow{BN})\\
=\frac{1}{3}\left[ -2.\frac{1}{2}(\overrightarrow{CA} +\overrightarrow{CB}) -4.\frac{1}{2}(\overrightarrow{BA} +\overrightarrow{BC})\right)\\
=\frac{1}{3}( -\overrightarrow{CA} -\overrightarrow{CB} -2\overrightarrow{BA} -2\overrightarrow{BC})\\
=\frac{1}{3}(\overrightarrow{AC} +\overrightarrow{BC} +2\overrightarrow{AB} +2\overrightarrow{CB})\\
=\frac{1}{3}(\overrightarrow{AC} +2\overrightarrow{AB} +\overrightarrow{CB})\\
=\frac{1}{3}(\overrightarrow{AB} +2\overrightarrow{AB}) =\frac{1}{3} .3\overrightarrow{AB} =\overrightarrow{AB} \ ( đpcm)
\end{array}$
b) Ta có:
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
-\frac{4}{3}\overrightarrow{CM} -\frac{2}{3}\overrightarrow{BN} =\frac{1}{3}( -4\overrightarrow{CM} -2\overrightarrow{BN})\\
=\frac{1}{3}\left[ -4.\frac{1}{2}(\overrightarrow{CA} +\overrightarrow{CB}) -2.\frac{1}{2}(\overrightarrow{BA} +\overrightarrow{BC})\right)\\
=\frac{1}{3}( -2\overrightarrow{CA} -2\overrightarrow{CB} -\overrightarrow{BA} -\overrightarrow{BC})\\
=\frac{1}{3}( 2\overrightarrow{AC} +2\overrightarrow{BC} +\overrightarrow{AB} +\overrightarrow{CB})\\
=\frac{1}{3}( 2\overrightarrow{AC} +\overrightarrow{AB} +\overrightarrow{BC})\\
=\frac{1}{3}( 2\overrightarrow{AC} +\overrightarrow{AC}) =\frac{1}{3} .3\overrightarrow{AC} =\overrightarrow{AC} \ ( đpcm)
\end{array}$
c) Ta có:
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\frac{1}{3}\overrightarrow{BN} -\frac{1}{3}\overrightarrow{CM} =\frac{1}{3}(\overrightarrow{BN} -\overrightarrow{CM})\\
=\frac{1}{3}\left[\frac{1}{2}(\overrightarrow{BA} +\overrightarrow{BC}) -\frac{1}{2}(\overrightarrow{CA} +\overrightarrow{CB})\right]\\
=\frac{1}{6}(\overrightarrow{BA} +\overrightarrow{BC} -\overrightarrow{CA} -\overrightarrow{CB})\\
=\frac{1}{6}(\overrightarrow{BA} +\overrightarrow{BC} +\overrightarrow{AC} +\overrightarrow{BC})\\
=\frac{1}{6}(\overrightarrow{BC} +2\overrightarrow{BC}) =\frac{1}{6} .3\overrightarrow{BC} =\frac{1}{2}\overrightarrow{BC}
\end{array}$
Mà M, N lần lượt là trung điểm của AB và AC trong tam giác ABC nên Mn là đường trung bình
Do đó $\displaystyle MN=\frac{1}{2} BC\Rightarrow \overrightarrow{MN} =\frac{1}{2}\overrightarrow{BC} \Rightarrow \overrightarrow{MN} =\frac{1}{3}\overrightarrow{BN} -\frac{1}{3}\overrightarrow{CM}$ (đpcm)