Câu 5:
\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}\)
\(=\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{BC}\)
\(=\overrightarrow{AB}-\dfrac{2}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\)
\(=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\)
\(c5\) \(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}=\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{BC}\)
\(=\overrightarrow{AB}+\dfrac{2}{3}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=\overrightarrow{AB}-\dfrac{2}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\)
\(=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\left(đpcm\right)\)
\(c4:\Rightarrow\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{GG'}\)
\(\overrightarrow{G'A'}+\overrightarrow{G'B'}+\overrightarrow{G'C'}=\overrightarrow{GG'}\)\(\Leftrightarrow\overrightarrow{G'A}+\overrightarrow{AA'}+\overrightarrow{G'B}+\overrightarrow{B'B}+\overrightarrow{G'C}+\overrightarrow{CC'}=\overrightarrow{GG'}\)
\(\Leftrightarrow\overrightarrow{AA'}+\overrightarrow{BB'}+\overrightarrow{CC'}=3\overrightarrow{GG'}\)\(\left(dpcm\right)\)
\(c3:a,\) \(2\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}=2\overrightarrow{IA}+2\overrightarrow{IM}=2\overrightarrow{MI}+2\left(\overrightarrow{IM}+\overrightarrow{MI}\right)=2.\overrightarrow{0}=\overrightarrow{0}\left(đpcm\right)\)
\(b,2\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\)
\(=2\left(\overrightarrow{OI}+\overrightarrow{IA}\right)+\overrightarrow{OI}+\overrightarrow{IB}+\overrightarrow{OI}+\overrightarrow{IC}\)
\(=4\overrightarrow{OI}+2\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}\)\(=4\overrightarrow{OI}+\overrightarrow{0}=4\overrightarrow{OI}\left(đpcm\right)\)
\(c2:\) \(\left\{{}\begin{matrix}3AH=2AB\\3AK=AC\\4BM=3MC\\\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\overrightarrow{AB}=\dfrac{3}{2}\overrightarrow{AH}\\\overrightarrow{AC}=3\overrightarrow{AK}\\\overrightarrow{BM}=\dfrac{3}{7}\overrightarrow{BC}\Rightarrow\overrightarrow{BC}=\dfrac{7}{3}\overrightarrow{BM}\end{matrix}\right.\)
\(\Rightarrow\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}=3\overrightarrow{AK}-\dfrac{3}{2}\overrightarrow{AH}\)
\(\Rightarrow\dfrac{7}{3}\overrightarrow{BM}=3\overrightarrow{AK}-\dfrac{3}{2}\overrightarrow{AH}\Leftrightarrow\overrightarrow{BM}=\dfrac{9}{7}\overrightarrow{AK}-\dfrac{9}{14}\overrightarrow{AH}\)
