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Tham khảo:
Dễ thấy: \(\overrightarrow {OA} = \overrightarrow {OM} + \overrightarrow {MA} \); \(\overrightarrow {OB} = \overrightarrow {OM} + \overrightarrow {MB} \)
Tương tự: \(\overrightarrow {OC} = \overrightarrow {ON} + \overrightarrow {NC} \); \(\overrightarrow {OD} = \overrightarrow {ON} + \overrightarrow {ND} \)
\(\begin{array}{l} \Rightarrow \overrightarrow {OA} + \overrightarrow {OB} + \overrightarrow {OC} + \overrightarrow {OD} = \left( {\overrightarrow {OM} + \overrightarrow {MA} } \right) + \left( {\overrightarrow {OM} + \overrightarrow {MB} } \right) + \left( {\overrightarrow {ON} + \overrightarrow {NC} } \right) + \left( {\overrightarrow {ON} + \overrightarrow {ND} } \right)\\ = \left( {\overrightarrow {OM} + \overrightarrow {OM} + \overrightarrow {MA} + \overrightarrow {MB} } \right) + \left( {\overrightarrow {ON} + \overrightarrow {ON} + \overrightarrow {NC} + \overrightarrow {ND} } \right)\\ = \overrightarrow {OM} + \overrightarrow {OM} + \overrightarrow {ON} + \overrightarrow {ON} \\ = \left( {\overrightarrow {OM} + \overrightarrow {ON} } \right) + \left( {\overrightarrow {OM} + \overrightarrow {ON} } \right)\\ = \overrightarrow 0 + \overrightarrow 0 \\ = \overrightarrow 0 .\end{array}\)
\(\overrightarrow{MN}=\overrightarrow{MC}+\overrightarrow{CN}=\dfrac{3}{4}\overrightarrow{AC}-\dfrac{1}{2}\overrightarrow{AB}=\dfrac{3}{4}\left(\overrightarrow{AB}+\overrightarrow{AD}\right)-\dfrac{1}{2}\overrightarrow{AB}\)
\(=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{3}{4}\overrightarrow{AD}\)
\(\Rightarrow a+b=\dfrac{1}{2}+\dfrac{3}{4}=...\)
dễ mà ,mình bỏ chữ vecto nha
IA+IB+IC+ID=IM+MA+IM+MB+IN+NC+IN+ND
=2IM+2IN+MA+MB+NC+ND
=0
Do M là trung điểm BC nên: \(\overrightarrow{AM}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\)
Tương tự: \(\overrightarrow{BN}=\dfrac{1}{2}\overrightarrow{BA}+\dfrac{1}{2}\overrightarrow{BC}\) ; \(\overrightarrow{CP}=\dfrac{1}{2}\overrightarrow{CA}+\dfrac{1}{2}\overrightarrow{CB}\)
Cộng vế:
\(\overrightarrow{AM}+\overrightarrow{BN}+\overrightarrow{CP}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{BA}+\dfrac{1}{2}\overrightarrow{BC}+\dfrac{1}{2}\overrightarrow{CA}+\dfrac{1}{2}\overrightarrow{CB}\)
\(=\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{BA}\right)+\dfrac{1}{2}\left(\overrightarrow{AC}+\overrightarrow{CA}\right)+\dfrac{1}{2}\left(\overrightarrow{BC}+\overrightarrow{CB}\right)=\overrightarrow{0}\)
b. Từ câu a ta có:
\(\overrightarrow{AM}+\overrightarrow{BN}+\overrightarrow{CP}=\overrightarrow{0}\)
\(\Leftrightarrow\overrightarrow{AO}+\overrightarrow{OM}+\overrightarrow{BO}+\overrightarrow{ON}+\overrightarrow{CO}+\overrightarrow{OP}=\overrightarrow{0}\)
\(\Leftrightarrow-\overrightarrow{OA}+\overrightarrow{OM}-\overrightarrow{OB}+\overrightarrow{ON}-\overrightarrow{OC}+\overrightarrow{OP}=\overrightarrow{0}\)
\(\Leftrightarrow\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{OM}+\overrightarrow{ON}+\overrightarrow{OP}\) (đpcm)
\(\overrightarrow{BM}+\overrightarrow{CN}+\overrightarrow{AP}\)
\(=\dfrac{1}{2}\left(\overrightarrow{BC}+\overrightarrow{CA}+\overrightarrow{AP}\right)\)
\(=\overrightarrow{0}\)
GIÚP MÌNH ĐI.........MÌNH ĐANG CẦN GẤP
Chứng minh:\(\overrightarrow{SA}+\overrightarrow{SB}+\overrightarrow{SC}+\overrightarrow{SD}=4\overrightarrow{SO}\)
Ta có: \(\overrightarrow{SA}+\overrightarrow{SB}+\overrightarrow{SC}+\overrightarrow{SD}=\overrightarrow{SO}+\overrightarrow{OA}+\overrightarrow{SO}+\overrightarrow{OB}+\overrightarrow{SO}+\overrightarrow{OC}+\overrightarrow{SO}+\overrightarrow{OD}\)\(=4\overrightarrow{SO}+\left(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}\right)\)
Mà: \(\overrightarrow{OA}+\overrightarrow{OB}=2\overrightarrow{OM}\) và \(\overrightarrow{OC}+\overrightarrow{OD}=2\overrightarrow{ON}\)
\(=4\overrightarrow{SO}+\left(2\overrightarrow{OM}+2\overrightarrow{ON}\right)\)
\(=4\overrightarrow{SO}+2\left(\overrightarrow{OM}+\overrightarrow{ON}\right)=4\overrightarrow{SO}+2.\overrightarrow{0}=4\overrightarrow{SO}\left(đpcm\right)\)