Lời giải:

a) Ta có:

\(\overrightarrow{AB}+\overrightarrow{CD}+\overrightarrow{BC}+\overrightarrow{DE}=(\overrightarrow{AB}+\overrightarrow{BC})+(\overrightarrow{CD}+\overrightarrow{DE})\)

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

\(\Rightarrow \overrightarrow{AB}+\overrightarrow{CD}=\overrightarrow{AE}-\overrightarrow{BC}-\overrightarrow{DE}\) (đpcm)

b)

\(\overrightarrow {AB}+\overrightarrow{DC}+\overrightarrow{BE}+\overrightarrow{ED}=(\overrightarrow{AB}+\overrightarrow{BE})+(\overrightarrow{ED}+\overrightarrow{DC})\)

\(=\overrightarrow{AE}+\overrightarrow{EC}=\overrightarrow{AC}\)

\(\Rightarrow \overrightarrow{AB}=\overrightarrow{AC}-\overrightarrow{DC}-\overrightarrow{BE}-\overrightarrow{ED}\) (đpcm)

Lời giải:

Ta có:

\(\overrightarrow{AB}+\overrightarrow{CD}=(\overrightarrow{AD}+\overrightarrow{DB})+(\overrightarrow{CB}+\overrightarrow{BD})\)

\(=(\overrightarrow{AD}+\overrightarrow{CB})+(\overrightarrow{DB}+\overrightarrow{BD})\)

\(=\overrightarrow{AD}+\overrightarrow{CB}\)

(\(\overrightarrow{DB}; \overrightarrow{BD}\) là 2 vector đối nhau nên tổng của chúng bằng vector 0)

Ta có đpcm

@Akai Haruma

Phương An

@Trần Việt Linh

Chuyển vế: \(\overrightarrow{AC}+\overrightarrow{BD}+\overrightarrow{EF}-\overrightarrow{AF}-\overrightarrow{BC}-\overrightarrow{ED}\)\(=\overrightarrow{AC}+\overrightarrow{BD}+\overrightarrow{EF}+\overrightarrow{FA}+\overrightarrow{CB}+\overrightarrow{DE}\)\(=\left(\overrightarrow{AC}+\overrightarrow{CB}\right)+\left(\overrightarrow{BD}+\overrightarrow{DE}\right)+\left(\overrightarrow{EF}+\overrightarrow{FA}\right)\)\(=\overrightarrow{AB}+\overrightarrow{BE}+\overrightarrow{EA}\)\(=\overrightarrow{AE}+\overrightarrow{EA}\)

\(=0\)

Suy ra: \(\overrightarrow{AC}+\overrightarrow{BD}+\overrightarrow{EF}=\overrightarrow{AF}+\overrightarrow{BC}+\overrightarrow{ED}\)

bài 1

a CO-OB=BA

<=.> CO = BA +OB

<=> CO=OA ( LUÔN ĐÚNG )=>ĐPCM

b AB-BC=DB

<=> AB=DB+BC

<=> AB=DC(LUÔN ĐÚNG )=> ĐPCM

Cc DA-DB=OD-OC

<=> DA+BD= OD+CO

<=> BA= CD (LUÔN ĐÚNG )=> ĐPCM

d DA-DB+DC=0

VT= DA +BD+DC

= BA+DC

Mà BA=CD(CMT)

=> VT= CD+DC=O

BÀI 2

AC=AB+BC

BD=BA+AD

=> AC+BD= AB+BC+BA+AD=BC+AD (đpcm)

b: \(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}+\overrightarrow{GD}\)

\(=2\overrightarrow{GE}+2\cdot\overrightarrow{GF}\)

\(=\overrightarrow{0}\)

a: 2 vecto IA+vecto IB=vecto 0

=>2 vecto IA=-vecto IB

=>I nằm giữa A và B và IA=2IB

=>vecto AI=2/3*vecto AB

b: 2/5vecto MA+3/5vecto MB

=2/5vecto MI+2/5vecto IA+3/5vecto MI+3/5vecto IB

=vetco MI+1/5(2 vecto IA+3 vecto IB)

=vecto MI