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\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}\)
\(=\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{BC}\)
\(=\overrightarrow{AB}+\dfrac{2}{3}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)\)
\(=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\)
A B C K I
a)
\(\overrightarrow{AK}=\overrightarrow{AI}+\overrightarrow{IK}=\overrightarrow{AI}+\dfrac{1}{2}\overrightarrow{IB}=\overrightarrow{AI}+\dfrac{1}{2}\left(\overrightarrow{IA}+\overrightarrow{AB}\right)\)
\(=\overrightarrow{AI}+\dfrac{1}{2}\overrightarrow{IA}+\dfrac{1}{2}\overrightarrow{AB}\)\(=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AI}\).
b) Theo câu a:
\(\overrightarrow{AK}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AI}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}.\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)
\(=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}=\dfrac{3}{4}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}\).
Ta có :
\(\overrightarrow{BP}+\overrightarrow{AN}+\overrightarrow{CM}=\overrightarrow{BC}+\overrightarrow{CP}+\overrightarrow{AB}+\overrightarrow{BN}+\overrightarrow{CA}+\overrightarrow{AM}=\overrightarrow{CP}+\overrightarrow{BN}+\overrightarrow{AM}\)\(=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{BC}+\dfrac{1}{3}\overrightarrow{CA}\)
\(=\dfrac{1}{3}\left(\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}\right)\)
\(=\dfrac{1}{3}\overrightarrow{0}\)
\(=\overrightarrow{0}\)
\(\RightarrowĐPCM\)
\(\overrightarrow{AD}=2\overrightarrow{DB}\Rightarrow\overrightarrow{AD}=\dfrac{2}{3}\overrightarrow{AB}\) ; \(\overrightarrow{CE}=3\overrightarrow{EA}\Rightarrow\overrightarrow{AE}=\dfrac{1}{4}\overrightarrow{AC}\)
Lại có M là trung điểm DE
\(\Rightarrow\overrightarrow{AM}=\dfrac{1}{2}\left(\overrightarrow{AD}+\overrightarrow{AE}\right)=\dfrac{1}{2}\left(\dfrac{2}{3}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}\right)=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{8}\overrightarrow{AC}\)
I là trung điểm BC \(\Rightarrow\overrightarrow{AI}=\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)
\(\Rightarrow\overrightarrow{MI}=\overrightarrow{MA}+\overrightarrow{AI}=\overrightarrow{AI}-\overrightarrow{AM}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}-\dfrac{1}{3}\overrightarrow{AB}-\dfrac{1}{8}\overrightarrow{AC}=\dfrac{1}{6}\overrightarrow{AB}+\dfrac{3}{8}\overrightarrow{AC}\)
Lời giải:
\(\overrightarrow{MN}=\overrightarrow{MB}+\overrightarrow{BN}=\overrightarrow{MB}+\overrightarrow{BC}+\overrightarrow{CN}\)
\(=\overrightarrow{MB}+\overrightarrow{BC}+2\overrightarrow{BC}=\overrightarrow{MB}+3\overrightarrow{BC}\)
\(=\overrightarrow{MA}+\overrightarrow{AB}+3(\overrightarrow{BA}+\overrightarrow{AC})\)
\(=-\overrightarrow{AM}+\overrightarrow{AB}-3\overrightarrow{AB}+3\overrightarrow{AC}\)
\(=-\frac{1}{3}\overrightarrow{AB}+\overrightarrow {AB}-3\overrightarrow{AB}+3\overrightarrow{AC}\)
\(=\frac{-7}{3}\overrightarrow{AB}+3\overrightarrow{AC}\)
Ta có đpcm.
Mình sửa lại câu hỏi: CM \(\overrightarrow{MN}=-\dfrac{7}{3}\overrightarrow{AB}+3\overrightarrow{AC}\)