Cho 6 điểm A , B , C , D , E , F chứng minh:
\(\overrightarrow{AB}+\overrightarrow{CD}+\overrightarrow{EF}=\overrightarrow{AD}+\overrightarrow{CF}+\overrightarrow{EB}\)
cho 6 điểm A, B , C , D , E , F bất kì trên mặt phẳng
chứng minh a, \(\overrightarrow{AB}+\overrightarrow{CD}=\overrightarrow{AD}+\overrightarrow{CB}\)
b , \(\overrightarrow{AB}+\overrightarrow{CD}+\overrightarrow{EA}=\overrightarrow{ED}+\overrightarrow{CB}\)
C, \(\overrightarrow{AD}+\overrightarrow{BE}+\overrightarrow{CF}=\overrightarrow{AE}+\overrightarrow{BF}+\overrightarrow{CD}=\overrightarrow{ÀF}+\overrightarrow{BD}+\overrightarrow{CE}\)
a.\(\overrightarrow{AB}+\overrightarrow{CD}=\overrightarrow{AD}+\overrightarrow{CB}\)
VT:\(\overrightarrow{AB}+\overrightarrow{CD}\)
=\(\overrightarrow{AC}+\overrightarrow{CB}+\overrightarrow{CA}+\overrightarrow{AD}\)
=\(\overrightarrow{AB}+\overrightarrow{CB}=0\left(đpcm\right)\)
b.\(\overrightarrow{AB}+\overrightarrow{CD}+\overrightarrow{EA}=\overrightarrow{ED}+\overrightarrow{CB}\)
\(\Leftrightarrow\overrightarrow{AB}+\overrightarrow{CD}+\overrightarrow{EA}+\overrightarrow{DE}+\overrightarrow{BC}=\overrightarrow{0}\)
\(\Leftrightarrow\overrightarrow{AC}+\overrightarrow{CE}+\overrightarrow{EA}=\overrightarrow{0}\)
\(\Leftrightarrow\overrightarrow{AE}+\overrightarrow{EA}=\overrightarrow{0}\)
\(\Leftrightarrow\overrightarrow{0}=\overrightarrow{0}\left(LĐ\right)\)
Ai có giải giúp mình câu này không:
Cho 6 điểm A, B, C, D, E, F. CMR:
\(a.\overrightarrow{CD}+\overrightarrow{FA}-\overrightarrow{BA}-\overrightarrow{ED}+\overrightarrow{BC}-\overrightarrow{FE}=\overrightarrow{0}\)
\(b.\overrightarrow{AD}-\overrightarrow{FC}-\overrightarrow{EB}=\overrightarrow{CD}-\overrightarrow{EA}-\overrightarrow{FB}\)
\(c.\overrightarrow{AB}-\overrightarrow{DC}-\overrightarrow{FE}=\overrightarrow{CF}-\overrightarrow{DA}+\overrightarrow{EB}\)
a, =CD+FA+AB+DE+BC+EF=(CD+DE)+(AB+BC)+FA+EF
=CE+AC+FA+EF= (CE+EF)+AC+FA=CF+AC+FA=(CF+FA)+AC=CA+AC=0
b,VP=CD+AE+BF
VT=AD+FC+BE=AC+CD+CB+BF+BA+AE=(AC+CB)+CD+BF+BA+AE
=AB+CD+BF+BA+AE=(AB+BA)+CD+BF+AE=CD+BF+AE=VP(dccm)
Cho 6 điểm A, B, C, D, E, F.CMR:
\(a.\overrightarrow{CD}+\overrightarrow{FA}-\overrightarrow{BA}-\overrightarrow{ED}+\overrightarrow{BC}-\overrightarrow{FE}=\overrightarrow{0}\)
\(b.\overrightarrow{AD}-\overrightarrow{FC}-\overrightarrow{EB}=\overrightarrow{CD}-\overrightarrow{EA}-\overrightarrow{FB}\)
\(c.\overrightarrow{AB}-\overrightarrow{DC}-\overrightarrow{FE}=\overrightarrow{CF}-\overrightarrow{DA}+\overrightarrow{EB}\)
Cho tứ giác ABCD. Gọi E, F, O lần lượt là trung điểm của AC, BD, EF. Chứng minh:
\(\overrightarrow{AB}+\overrightarrow{CD}=\overrightarrow{AD}+\overrightarrow{CB}\)
\(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=\overrightarrow{0}\)
\(\overrightarrow{AB}+\overrightarrow{CD}=\overrightarrow{AB}+\overrightarrow{CB}+\overrightarrow{BD}=\overrightarrow{AB}+\overrightarrow{BD}+\overrightarrow{CB}=\overrightarrow{AD}+\overrightarrow{CB}\)
\(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=\left(\overrightarrow{OE}+\overrightarrow{EA}\right)+\left(\overrightarrow{OF}+\overrightarrow{FB}\right)+\left(\overrightarrow{OE}+\overrightarrow{EC}\right)+\left(\overrightarrow{OF}+\overrightarrow{FD}\right)\)
\(=2\left(\overrightarrow{OE}+\overrightarrow{EF}\right)+\left(\overrightarrow{EA}+\overrightarrow{EC}\right)+\left(\overrightarrow{FB}+\overrightarrow{FD}\right)\)
\(=2.\overrightarrow{0}+\overrightarrow{0}+\overrightarrow{0}=\overrightarrow{0}\)
Cho 4 điểm A,B,C,D. Gọi E,F,G lần lượt là trung điểm của AB,CD,EF. Chứng minh
a,\(\overrightarrow{AC}+\overrightarrow{BD}=2\overrightarrow{EF}\)
b,\(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}+\overrightarrow{GD}=\overrightarrow{0}\)
c,\(\overrightarrow{AB}+\overrightarrow{AC+}\overrightarrow{AD}=4\overrightarrow{AG}\)
MÌNH CẦN GẤP GIÚP MÌNH NHA
Cho 6 điểm A, B, C, D, E, F. chứng minh các đẳng thức vecto sau:
a) \(\overrightarrow{AB}-\overrightarrow{CD}=\overrightarrow{AC}-\overrightarrow{BD}\)
b) \(\overrightarrow{AB}+\overrightarrow{DC}+\overrightarrow{BD}+\overrightarrow{CA}=\overrightarrow{0}\)
c) \(\overrightarrow{AC}+\overrightarrow{DE}-\overrightarrow{DC}-\overrightarrow{CE}+\overrightarrow{CB}=\overrightarrow{AB}\)
d) \(\overrightarrow{AB}+\overrightarrow{DE}+\overrightarrow{CF}=\overrightarrow{AC}+\overrightarrow{DF}+\overrightarrow{CB}+\overrightarrow{CE}\)
HELP ME!!
\(a\text{) }\overrightarrow{AB}-\overrightarrow{CD}=\left(\overrightarrow{AC}+\overrightarrow{CB}\right)-\overrightarrow{CD}\\ =\overrightarrow{AC}-\left(\overrightarrow{CD}-\overrightarrow{CB}\right)=\overrightarrow{AC}-\overrightarrow{BD}\)
\(b\text{) }\overrightarrow{AB}+\overrightarrow{DC}+\overrightarrow{BD}+\overrightarrow{CA}=\left(\overrightarrow{AB}+\overrightarrow{BD}\right)+\left(\overrightarrow{DC}+\overrightarrow{CA}\right)\\ =\left(\overrightarrow{AB}+\overrightarrow{BD}\right)+\left(\overrightarrow{DC}+\overrightarrow{CA}\right)=\overrightarrow{AD}+\overrightarrow{DA}=0\)
\(c\text{) }\overrightarrow{AC}+\overrightarrow{DE}-\overrightarrow{DC}-\overrightarrow{CE}+\overrightarrow{CB}\\ =\left(\overrightarrow{AC}+\overrightarrow{CB}\right)+\left(\overrightarrow{DE}-\overrightarrow{DC}\right)-\overrightarrow{CE}\\ =\overrightarrow{AB}+\overrightarrow{CE}-\overrightarrow{CE}=\overrightarrow{AB}\)
\(d\text{) }\overrightarrow{AB}+\overrightarrow{DE}+\overrightarrow{CF}\\ =\left(\overrightarrow{AC}+\overrightarrow{CB}\right)+\left(\overrightarrow{DF}+\overrightarrow{FE}\right)+\left(\overrightarrow{CE}+\overrightarrow{EF}\right)\\ =\overrightarrow{AC}+\overrightarrow{CE}+\overrightarrow{CB}+\overrightarrow{DF}+\left(\overrightarrow{FE}+\overrightarrow{EF}\right)\\ =\overrightarrow{AC}+\overrightarrow{CE}+\overrightarrow{CB}+\overrightarrow{DF}\)
cho Tứ giác ABCD có E, F là trung điểm AB, CD. O là trung điểm của EF
a) Chứng minh \(\overrightarrow{AD}+\overrightarrow{BC}=2\overrightarrow{EF}\)
b) Chứng minh \(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=\overrightarrow{0}\)
c) Chứng minh \(\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}+\overrightarrow{MD}=4\overrightarrow{MO}\)
d) Xác định M để \(\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}+\overrightarrow{MD}\) nhỏ nhất
Cho bốn điểm \(A, B, C, D\). Chứng minh rằng:
a) \(\overrightarrow {AB} + \overrightarrow {BC} + \overrightarrow {CD} + \overrightarrow {DA} = \overrightarrow 0 \)
b) \(\overrightarrow {AC} - \overrightarrow {AD} = \overrightarrow {BC} - \overrightarrow {BD} \)
a)
\(\begin{array}{l}\overrightarrow {AB} + \overrightarrow {BC} + \overrightarrow {CD} + \overrightarrow {DA} = \left( {\overrightarrow {AB} + \overrightarrow {BC} } \right) + \left( {\overrightarrow {CD} + \overrightarrow {DA} } \right)\\ = \overrightarrow {AC} + \overrightarrow {CA} = \overrightarrow {AA} = \overrightarrow 0 .\end{array}\)
b)
\(\overrightarrow {AC} - \overrightarrow {AD} = \overrightarrow {DC} \) và \(\overrightarrow {BC} - \overrightarrow {BD} = \overrightarrow {DC} \)
\( \Rightarrow \overrightarrow {AC} - \overrightarrow {AD} = \overrightarrow {BC} - \overrightarrow {BD} \)
Cho bốn điểm A, B, C, D. Chứng minh:
a) \(\overrightarrow {AB} + \overrightarrow {CD} = \overrightarrow {AD} + \overrightarrow {CB} \)
b) \(\overrightarrow {AB} + \overrightarrow {CD} + \overrightarrow {BC} + \overrightarrow {DA} = \overrightarrow 0 \)
a)
\(\begin{array}{l}\overrightarrow {AB} + \overrightarrow {CD} = \overrightarrow {AD} + \overrightarrow {CB} \\ \Leftrightarrow \overrightarrow {AB} - \overrightarrow {CB} = \overrightarrow {AD} - \overrightarrow {CD} \\ \Leftrightarrow \overrightarrow {AB} + \overrightarrow {BC} = \overrightarrow {AD} + \overrightarrow {DC} \\ \Leftrightarrow \overrightarrow {AC} = \overrightarrow {AC} \end{array}\)
(luôn đúng)
b) \(\overrightarrow {AB} + \overrightarrow {CD} + \overrightarrow {BC} + \overrightarrow {DA} = \overrightarrow 0 \)
Ta có:
\(\begin{array}{l}\overrightarrow {AB} + \overrightarrow {CD} + \overrightarrow {BC} + \overrightarrow {DA} = (\overrightarrow {AB} + \overrightarrow {BC} ) + (\overrightarrow {CD} + \overrightarrow {DA} )\\ = \overrightarrow {AC} + \overrightarrow {CA} = \overrightarrow 0 \end{array}\)
Chú ý khi giải
+) Hiệu hai vecto chung gốc: \(\overrightarrow {AB} - \overrightarrow {AC} = \overrightarrow {CB} \) (suy ra từ tổng \(\overrightarrow {AB} = \overrightarrow {AC} + \overrightarrow {CB} \))
+) Với 4 điểm A, B, C, D bất kì ta có: \(\overrightarrow {AB} + \overrightarrow {BC} + \overrightarrow {CD} + \overrightarrow {DA} = \overrightarrow {AA} = \overrightarrow 0 \)