A. \(\overrightarrow{AB}=\frac{2}{3}\overrightarrow{AN}+\frac{2}{3}\overrightarrow{CM}\).
B. \(\overrightarrow{AB}=\frac{4}{3}\overrightarrow{AN}-\frac{2}{3}\overrightarrow{CM}\)
C. \(\overrightarrow{AB}=\frac{4}{3}\overrightarrow{AN}+\frac{4}{3}\overrightarrow{CM}\).
D. \(\overrightarrow{AB}=\frac{4}{3}\overrightarrow{AN}+\frac{2}{3}\overrightarrow{CM}\).
D
Ta có \(\overrightarrow{AN}=\frac{1}{2}\left( \overrightarrow{AB}+\overrightarrow{AC} \right)=\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AC}\)
\(\overrightarrow{CM}=\overrightarrow{CA}+\overrightarrow{AM}\Rightarrow \frac{1}{2}\overrightarrow{CM}=\frac{1}{2}\overrightarrow{CA}+\frac{1}{2}\overrightarrow{AM}\)
Suy ra \(\overrightarrow{AN}+\frac{1}{2}\overrightarrow{CM}=\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AC}+\frac{1}{2}\overrightarrow{CA}+\frac{1}{2}\overrightarrow{AM}=\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AC}-\frac{1}{2}\overrightarrow{AC}+\frac{1}{2}\cdot \frac{1}{2}\overrightarrow{AB}=\frac{3}{4}\overrightarrow{AB}\)
Do đó \(\overrightarrow{AB}=\frac{4}{3}\overrightarrow{AN}+\frac{2}{3}\overrightarrow{CM}\).
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