a: \(\overrightarrow{AE}=\dfrac{2}{3}\overrightarrow{EC}\)

=>E nằm giữa A và C và AE=2/3EC

Ta có: AE+EC=AC(E nằm giữa A và C)

=>\(AC=\dfrac{2}{3}EC+EC=\dfrac{5}{3}EC\)

=>\(\dfrac{AE}{AC}=\dfrac{\dfrac{2}{3}EC}{\dfrac{5}{3}EC}=\dfrac{2}{3}:\dfrac{5}{3}=\dfrac{2}{5}\)

=>\(AE=\dfrac{2}{5}AC\)

=>\(\overrightarrow{AE}=\dfrac{2}{5}\cdot\overrightarrow{AC}\)

\(\overrightarrow{BE}=\overrightarrow{BA}+\overrightarrow{AE}\)

\(=-\overrightarrow{AB}+\dfrac{2}{5}\cdot\overrightarrow{AC}\)

b: \(\left|\overrightarrow{IA}+\overrightarrow{IG}\right|=\left|\overrightarrow{IA}-\overrightarrow{IG}\right|\)

=>\(\left[{}\begin{matrix}\overrightarrow{IA}+\overrightarrow{IG}=\overrightarrow{IA}-\overrightarrow{IG}\\\overrightarrow{IA}+\overrightarrow{IG}=\overrightarrow{IG}-\overrightarrow{IA}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2\cdot\overrightarrow{IG}=\overrightarrow{0}\\2\cdot\overrightarrow{IA}=\overrightarrow{0}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}I\equiv G\\I\equiv A\end{matrix}\right.\)

A B C M N K P

a) \(\overrightarrow{BC}=\overrightarrow{BA}+\overrightarrow{AC}=-2\overrightarrow{AM}+\frac{3}{2}\overrightarrow{AN}\)

b) Kẻ hình bình hành AMPN, ta có:

\(\overrightarrow{AK}=\frac{1}{2}\overrightarrow{AP}=\frac{1}{2}\left(\overrightarrow{AM}+\overrightarrow{AN}\right)=\frac{1}{2}\left(\frac{1}{2}\overrightarrow{AB}+\frac{2}{3}\overrightarrow{AC}\right)=\frac{1}{4}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\)

a: 

b: \(\overrightarrow{MN}=\overrightarrow{MA}+\overrightarrow{AN}\)

\(=\overrightarrow{CB}+\dfrac{1}{2}\cdot\overrightarrow{AK}\)

\(=\overrightarrow{CA}+\overrightarrow{AB}+\dfrac{1}{2}\cdot\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)

\(=-\overrightarrow{AC}+\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}\)

\(=\dfrac{5}{4}\cdot\overrightarrow{AB}-\dfrac{3}{4}\cdot\overrightarrow{AC}\)

Xét ΔMDC có N là trung điểm của DC

nên \(2\cdot\overrightarrow{MN}=\overrightarrow{MD}+\overrightarrow{MC}=\overrightarrow{MA}+\overrightarrow{AD}+\overrightarrow{MB}+\overrightarrow{BC}=\overrightarrow{AD}+\overrightarrow{BC}\)

Câu 1: 

Gọi M là trung điểm của AC

AM=AC/2=2

\(BM=\sqrt{3^2+2^2}=\sqrt{13}\)

\(\left|\overrightarrow{AB}+\overrightarrow{CB}\right|=\left|\overrightarrow{BA}+\overrightarrow{BC}\right|=2\cdot BM=2\sqrt{13}\)

Câu 6:

\(\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DE}+\overrightarrow{EF}+\overrightarrow{FA}\)

\(=\overrightarrow{AC}+\overrightarrow{CE}+\overrightarrow{EA}=\overrightarrow{AE}+\overrightarrow{EA}=\overrightarrow{0}\)

Ta có: AP+PC=AC

=>\(AC=\frac12PC+PC=\frac32PC\)

=>\(AP=\frac13AC\)

TA có: \(AM=\frac12\times AB\)

=>\(S_{CMA}=\frac12\times S_{CAB}=\frac12\times180=90\left(dm^2\right)\)

Ta có: \(AP=\frac13\times AC\)

=>\(S_{APM}=\frac13\times S_{AMC}=\frac13\times90=30\left(dm^2\right)\)

Ta có: \(CN=\frac12\times CB\)

=>\(S_{ANC}=\frac12\times S_{ACB}=\frac12\times180=90\left(dm^2\right)\)

Ta có: \(CP=\frac23\times CA\)

=>\(S_{CPN}=\frac23\times S_{CNA}=\frac23\times90=60\left(dm^2\right)\)

Ta có: \(BN=\frac12\times BC\)

=>\(S_{ANB}=\frac12\times S_{ABC}=\frac12\times180=90\left(dm^2\right)\)

Ta có: \(BM=\frac12\times BA\)

=>\(S_{NBM}=\frac12\times S_{ANB}=\frac12\times90=45\left(dm^2\right)\)

Ta có: \(S_{APM}+S_{NBM}+S_{MPCN}=S_{ABC}\)

=>\(S_{MPCN}=180-30-45=180-75=105\left(dm^2\right)\)