a) trong lục giác đều ABCDEF:
\(\overrightarrow{AD}=2\overrightarrow{AB}+2\overrightarrow{AF}\)
b Góc giữa \(\overrightarrow{AB}\) và \(\overrightarrow{BC}\) là \(60^{\degree}\)
\(\left\vert\frac12\overrightarrow{AB}+\frac12\overrightarrow{BC}\right\vert=\frac{\sqrt3}{2}a\) \(\left\vert\frac12\overrightarrow{AB}+\frac12\overrightarrow{BC}\right\vert=\frac12\sqrt{a^2+a^2+2a^2\cos60\degree}=\frac{\sqrt3}{2}a\)
a) \(\overrightarrow{AD}=2\overrightarrow{AB}+2\overrightarrow{AF}\)
b)\(\left\vert\frac12\overrightarrow{AB}+\frac12\overrightarrow{BC}\right\vert=\frac{\sqrt3}{2}a\)

\(M\) là trung điểm \(AB\rarr\overrightarrow{AM}\frac12\overrightarrow{b}\)
\(NA=2NC\rarr\overrightarrow{AN}=\frac23\overrightarrow{c}\)
\(K\) trung điểm \(MN\rarr\overrightarrow{AK}=\frac12(\overrightarrow{AM}+\overrightarrow{AN})=\frac14\overrightarrow{b}+\frac13\overrightarrow{c}\)
\(\overrightarrow{AK}=\frac14\overrightarrow{AB}+\frac13\overrightarrow{AC}\)

\(M\) trung điểm\(AB\rarr\overrightarrow{AM}=\frac12\overrightarrow{b}\)
\(NA=2NC\rarr\overrightarrow{AN}=\frac23\overrightarrow{c}\)
\(K\) trung điểm \(MN\rarr\overrightarrow{AK}=\frac12(\overrightarrow{AM}+\overrightarrow{AN})=\frac14\overrightarrow{b}+\frac13\overrightarrow{c}\)
\(\overrightarrow{AK}=\frac14\overrightarrow{AB}_{}+\frac13\overrightarrow{AC}\)