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\(3\overrightarrow{BI}=2\overrightarrow{IC}\Rightarrow3\overrightarrow{BI}=2\overrightarrow{IB}+2\overrightarrow{BC}\Rightarrow\overrightarrow{BI}=\frac{2}{5}\overrightarrow{BC}\)
\(5\overrightarrow{JB}=2\overrightarrow{JC}\Leftrightarrow5\overrightarrow{JB}=2\overrightarrow{JB}+2\overrightarrow{BC}\Rightarrow\overrightarrow{JB}=\frac{2}{3}\overrightarrow{BC}\)
\(\overrightarrow{AI}=\overrightarrow{AB}+\overrightarrow{BI}=\overrightarrow{AB}+\frac{2}{5}\overrightarrow{BC}=\overrightarrow{AB}+\frac{2}{5}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=\frac{3}{5}\overrightarrow{AB}+\frac{2}{5}\overrightarrow{AC}\)
\(\overrightarrow{AJ}=\overrightarrow{AB}+\overrightarrow{BJ}=\overrightarrow{AB}-\frac{2}{3}\overrightarrow{BC}=\overrightarrow{AB}-\frac{2}{3}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=\frac{5}{3}\overrightarrow{AB}-\frac{2}{3}\overrightarrow{AC}\)
\(\left\{{}\begin{matrix}2\overrightarrow{CI}=-3\overrightarrow{BI}\\5\overrightarrow{JB}=2\overrightarrow{JC}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2\overrightarrow{CB}+2\overrightarrow{BI}=-3\overrightarrow{BI}\\5\overrightarrow{JB}=2\overrightarrow{JB}+2\overrightarrow{BC}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\overrightarrow{BI}=-\frac{2}{5}\overrightarrow{BC}\\\overrightarrow{JB}=\frac{2}{3}\overrightarrow{BC}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\overrightarrow{AI}=\overrightarrow{AB}+\overrightarrow{BI}=\overrightarrow{AB}-\frac{2}{5}\overrightarrow{BC}\\\overrightarrow{AJ}=\overrightarrow{AB}+\overrightarrow{BJ}=\overrightarrow{AB}-\frac{2}{3}\overrightarrow{BC}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AI}=\overrightarrow{AB}-\frac{2}{5}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=\frac{7}{5}\overrightarrow{AB}-\frac{2}{5}\overrightarrow{AC}\\\overrightarrow{AJ}=\overrightarrow{AB}-\frac{2}{3}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=\frac{5}{3}\overrightarrow{AB}-\frac{2}{3}\overrightarrow{AC}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}7\overrightarrow{AB}-2\overrightarrow{AC}=5\overrightarrow{AI}\\5\overrightarrow{AB}-2\overrightarrow{AC}=3\overrightarrow{AJ}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AB}=\frac{5}{2}\overrightarrow{AI}-\frac{3}{2}\overrightarrow{AJ}\\\overrightarrow{AC}=\frac{25}{4}\overrightarrow{AI}-\frac{21}{4}\overrightarrow{AJ}\end{matrix}\right.\)
\(\overrightarrow{AG}=\frac{1}{3}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)=\frac{1}{3}\left(\frac{5}{2}\overrightarrow{AI}-\frac{3}{2}\overrightarrow{AJ}+\frac{25}{4}\overrightarrow{AI}-\frac{21}{4}\overrightarrow{AJ}\right)=...\)
Mình đang cần cách giải bài này mà không cần dựa vào vecto AB, AC á bạn
A B C D I M
a)
\(\overrightarrow{AI}=\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AD}\right)=\dfrac{1}{2}\left(\overrightarrow{AB}+\dfrac{3}{4}\overrightarrow{AC}\right)=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{3}{8}\overrightarrow{AC}\).
b)
\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}=\overrightarrow{AB}+x\overrightarrow{BC}\)\(=\overrightarrow{AB}+x\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=\left(1-x\right)\overrightarrow{AB}+x\overrightarrow{AC}\).
c) A, M, I thẳng hàng khi và chỉ khi hai véc tơ \(\overrightarrow{AM};\overrightarrow{AI}\) cùng phương
hay \(\dfrac{1-x}{\dfrac{1}{2}}=\dfrac{x}{\dfrac{3}{8}}\Leftrightarrow\dfrac{3}{8}\left(1-x\right)=\dfrac{1}{2}x\)
\(\Leftrightarrow\dfrac{7}{8}x=\dfrac{3}{8}\)\(\Leftrightarrow x=\dfrac{3}{7}\).
