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1) Ta có:\(\overrightarrow{AB}+\overrightarrow{DE}-\overrightarrow{DB}+\overrightarrow{BC}=\overrightarrow{AE}+\overrightarrow{BC}=\overrightarrow{AC}+\overrightarrow{CE}+\overrightarrow{BE}+\overrightarrow{EC}\)
\(=\overrightarrow{AC}+\overrightarrow{BE}+\overrightarrow{CE}+\overrightarrow{EC}=\overrightarrow{AC}+\overrightarrow{BE}\left(đpcm\right)\)2) a) Ta có: \(\overrightarrow{AD}+\overrightarrow{BE}+\overrightarrow{CF}=\overrightarrow{AE}+\overrightarrow{ED}+\overrightarrow{BF}+\overrightarrow{FE}+\overrightarrow{CD}+\overrightarrow{DF}\)\(=\overrightarrow{AE}+\overrightarrow{BF}+\overrightarrow{CD}+\overrightarrow{ED}+\overrightarrow{DF}+\overrightarrow{FE}\)
\(=\overrightarrow{AE}+\overrightarrow{BF}+\overrightarrow{CD}\left(đpcm\right)\)
b) Ta có: \(\overrightarrow{AB}+\overrightarrow{CD}=\overrightarrow{AD}+\overrightarrow{DB}+\overrightarrow{CB}+\overrightarrow{BD}\)
\(=\overrightarrow{AD}+\overrightarrow{CB}+\overrightarrow{DB}+\overrightarrow{BD}=\overrightarrow{AD}+\overrightarrow{CB}\left(đpcm\right)\)c) \(\overrightarrow{AB}-\overrightarrow{CD}=\overrightarrow{AB}-\overrightarrow{BD}\)
\(\overrightarrow{AB}+\overrightarrow{DC}=\overrightarrow{AB}+\overrightarrow{DB}\)
Ta có: \(\overrightarrow{AB}+\overrightarrow{DC}=\overrightarrow{AB}+\overrightarrow{DB}+\overrightarrow{BC}\) ( đề bài bị lỗi gì à ?? :v ) hay do mình =))
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
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)
a/ Theo quy tắc 3 điểm: \(\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}\)
\(\overrightarrow{AD}=\overrightarrow{AO}+\overrightarrow{OD}\)
\(\Rightarrow\overrightarrow{AD}+\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}+\overrightarrow{AO}+\overrightarrow{OD}\)
\(\overrightarrow{OD}=-\overrightarrow{OB}\)
\(\Rightarrow\overrightarrow{AD}+\overrightarrow{AB}=2\overrightarrow{AO}\)
b/ \(\overrightarrow{AC}=2\overrightarrow{AO}=2\overrightarrow{a};\overrightarrow{BD}=2\overrightarrow{BO}=2\overrightarrow{b}\)
\(\overrightarrow{BC}=\overrightarrow{BO}+\overrightarrow{OC}=\overrightarrow{BO}+\overrightarrow{AO}=\overrightarrow{a}+\overrightarrow{b}=-\overrightarrow{DA}\)
\(\overrightarrow{AB}=-\overrightarrow{CD}=\overrightarrow{AO}+\overrightarrow{OB}=\overrightarrow{a}-\overrightarrow{b}\)