a) \(2\overrightarrow{JA}+5\overrightarrow{JB}+3\overrightarrow{JC}=\overrightarrow{0}\)
\(\Leftrightarrow2\overrightarrow{JA}+2\overrightarrow{JB}+3\overrightarrow{JB}+3\overrightarrow{JC}=\overrightarrow{0}\)
\(\Leftrightarrow2\left(\overrightarrow{JA}+\overrightarrow{JB}\right)+3\left(\overrightarrow{JB}+\overrightarrow{JC}\right)=\overrightarrow{0}\)
\(\Leftrightarrow4\overrightarrow{JM}+6\overrightarrow{JN}=\overrightarrow{0}\)
\(\Leftrightarrow\overrightarrow{JM}=-\dfrac{3}{2}\overrightarrow{JN}\)
\(\Rightarrow M;N:J\) thẳng hàng \(\Rightarrow\) Đúng
b) Sai (Chứng minh trên)
c) \(2\overrightarrow{IA}+3\overrightarrow{IC}=\overrightarrow{0}\)
\(\Leftrightarrow2\overrightarrow{IJ}+2\overrightarrow{JA}+3\overrightarrow{IJ}+3\overrightarrow{JC}=\overrightarrow{0}\)
\(\Leftrightarrow5\overrightarrow{IJ}+2\overrightarrow{JA}+3\overrightarrow{JC}=\overrightarrow{0}\)
mà \(2\overrightarrow{JA}+5\overrightarrow{JB}+3\overrightarrow{JC}=\overrightarrow{0}\)
\(\Rightarrow5\overrightarrow{IJ}-5\overrightarrow{JB}=\overrightarrow{0}\) (trừ vế với vế)
\(\Leftrightarrow\overrightarrow{IJ}=\overrightarrow{JB}\)
\(\Rightarrow J\) là trung điểm \(BI\Rightarrow\) Đúng
d) \(\overrightarrow{AE}=\dfrac{5}{7}\overrightarrow{AB}\)
\(\Leftrightarrow\overrightarrow{AJ}+\overrightarrow{JE}=\dfrac{5}{7}\overrightarrow{AJ}+\dfrac{5}{7}\overrightarrow{JB}\)
\(\Leftrightarrow\overrightarrow{JE}=-\dfrac{2}{7}\overrightarrow{AJ}+\dfrac{5}{7}\overrightarrow{JB}=\dfrac{2}{7}\overrightarrow{JA}+\dfrac{5}{7}\overrightarrow{JB}\left(1\right)\)
Ta lại có \(2\overrightarrow{JA}+5\overrightarrow{JB}+3\overrightarrow{JC}=\overrightarrow{0}\)
\(\Leftrightarrow\overrightarrow{JC}=-\dfrac{2}{3}\overrightarrow{JA}-\dfrac{5}{3}\overrightarrow{JB}\left(2\right)\)
\(\left(1\right);\left(2\right)\) Ta thấy \(\dfrac{\dfrac{2}{7}}{-\dfrac{2}{3}}=\dfrac{\dfrac{5}{7}}{-\dfrac{5}{3}}=-\dfrac{3}{7}\)
\(\Rightarrow\overrightarrow{JE}=-\dfrac{3}{7}\overrightarrow{JC}\)
\(\Rightarrow C,E,J\) thẳng hàng \(\Rightarrow\) Đúng
