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1/ \(\overrightarrow{AM}=3\overrightarrow{AM}+\overrightarrow{MB}+\overrightarrow{MC}+\overrightarrow{MD}\)

\(\Leftrightarrow2\overrightarrow{AM}+3\overrightarrow{MG}=\overrightarrow{0}\)

\(\Leftrightarrow2\overrightarrow{AM}+3\overrightarrow{MA}+3\overrightarrow{AG}=\overrightarrow{0}\)

\(\Leftrightarrow\overrightarrow{AM}=3\overrightarrow{AG}\)

Ban tu ket luan

2/ Bạn coi lại đề bài, đẳng thức kia có vấn đề. 2k-1IB??

c) \(\overrightarrow{BG}+\overrightarrow{GC}=\overrightarrow{BC}\ne\overrightarrow{GA}\)

d) \(\overrightarrow{GB}+\overrightarrow{GC}=\dfrac{1}{2}\overrightarrow{GM}\ne\overrightarrow{GM}\)

a: vecto DE

=vecto DA+vecto AE

=-2vecto AB+2/5*vecto AC

vecto DG=vecto DB+vecto BG

=-2*vecto AB-vecto GB

=-2vecto AB-(-vecto GA-vecto GC)

=-2 vecto AB-(vecto CG-vecto GA)

=-2vecto AB-(vecto CG+vecto AG)

=-2vecto AB+vecto GA+vecto GC

=-2*vecto AB+2*vecto GF

=-2vecto AB+2*1/3*vecto BF

=-2*vecto AB+2/3(vecto BA+vecto BC)

=-2vecto AB-2/3vecto AB+2/3*veto BC

=-8/3vecto AB+2/3*(vecto BA+vecto AC)

=-10/3vecto AB+2/3vecto AC

b: vecto DE=-2vecto AB+2/5vecto AC

vecto DG=-10/3vecto AB+2/3*vecto AC

Vì \(\dfrac{-2}{-\dfrac{10}{3}}=2:\dfrac{10}{3}=\dfrac{6}{10}=\dfrac{3}{5}=\dfrac{2}{5}:\dfrac{2}{3}\)

nên D,E,G thẳng hàng

Do G là trọng tâm ABC \(\Rightarrow\overrightarrow{BG}=\dfrac{1}{3}\overrightarrow{BA}+\dfrac{1}{3}\overrightarrow{BC}\)

I đối xứng B qua G \(\Rightarrow\) \(\overrightarrow{BI}=2\overrightarrow{BG}=\dfrac{2}{3}\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{BC}=\dfrac{2}{3}\overrightarrow{BA}+\dfrac{2}{3}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)\)

\(\Rightarrow\overrightarrow{BI}=\dfrac{4}{3}\overrightarrow{BA}+\dfrac{2}{3}\overrightarrow{AC}=-\dfrac{4}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\)

\(\Rightarrow\overrightarrow{CI}=\overrightarrow{CB}+\overrightarrow{BI}=\overrightarrow{CA}+\overrightarrow{AB}-\dfrac{4}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\)

\(\Rightarrow\overrightarrow{CI}=-\dfrac{1}{3}\overrightarrow{AB}-\dfrac{1}{3}\overrightarrow{AC}\)

\(\overrightarrow{DE}=\overrightarrow{DA}+\overrightarrow{AE}=-2\overrightarrow{AB}+\frac{2}{5}\overrightarrow{AC}\)

\(\overrightarrow{DG}=\overrightarrow{DA}+\overrightarrow{AG}=-2\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}=-\frac{5}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}=\frac{5}{6}\left(-2\overrightarrow{AB}+\frac{2}{5}\overrightarrow{AC}\right)\)

\(\Rightarrow\overrightarrow{DG}=\frac{5}{6}\overrightarrow{DE}\Rightarrow\overrightarrow{DE}=\frac{6}{5}\overrightarrow{DG}\Rightarrow x=\frac{6}{5}\)