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a;\(\overrightarrow{AB}+2\overrightarrow{AC}\)
\(=\overrightarrow{AM}+\overrightarrow{MB}+2\overrightarrow{AM}+2\overrightarrow{MC}\)
\(=3\overrightarrow{AM}\)
b: \(\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}\)
\(=\overrightarrow{MG}+\overrightarrow{GA}+\overrightarrow{MG}+\overrightarrow{GB}+\overrightarrow{MG}+\overrightarrow{GC}\)
=3vecto MG
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.
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:
\(\overrightarrow{AK}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AI}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}.\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)
\(=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}=\dfrac{3}{4}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}\).
\(\overrightarrow{AD}=2\overrightarrow{DB}\Rightarrow\overrightarrow{AD}=\dfrac{2}{3}\overrightarrow{AB}\) ; \(\overrightarrow{CE}=3\overrightarrow{EA}\Rightarrow\overrightarrow{AE}=\dfrac{1}{4}\overrightarrow{AC}\)
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 \(\Rightarrow\overrightarrow{AI}=\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)
\(\Rightarrow\overrightarrow{MI}=\overrightarrow{MA}+\overrightarrow{AI}=\overrightarrow{AI}-\overrightarrow{AM}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}-\dfrac{1}{3}\overrightarrow{AB}-\dfrac{1}{8}\overrightarrow{AC}=\dfrac{1}{6}\overrightarrow{AB}+\dfrac{3}{8}\overrightarrow{AC}\)
Ta có: \(\overrightarrow{MB}=3\overrightarrow{MC}\Rightarrow\overrightarrow{MB}=3\left(\overrightarrow{MB}+\overrightarrow{BC}\right)\)
\(\Rightarrow\overrightarrow{MB}=3\overrightarrow{MB}+3\overrightarrow{BC}\)
\(\Rightarrow-\overrightarrow{MB}=3\overrightarrow{BC}\)
\(\Rightarrow\overrightarrow{BM}=\dfrac{2}{3}\overrightarrow{BC}\). Mà \(\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}\) nên \(\overrightarrow{BM}=\dfrac{2}{3}\left(\overrightarrow{AC}-\overrightarrow{AB}\right)\)
Theo quy tắc 3 điểm, ta có
\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}\Rightarrow\overrightarrow{AM}=\overrightarrow{AB}+\dfrac{3}{2}\overrightarrow{AC}-\dfrac{3}{2}\overrightarrow{AB}\)
\(\Rightarrow\overrightarrow{AM}=-\dfrac{1}{2}\overrightarrow{AB}+\dfrac{3}{2}\overrightarrow{AC}\) hay \(\overrightarrow{AM}=-\dfrac{1}{2}\overrightarrow{u}+\dfrac{3}{2}\overrightarrow{v}\)
\(\overrightarrow{MB}=-2\overrightarrow{MC}\Leftrightarrow\overrightarrow{MB}=-2\left(\overrightarrow{MB}+\overrightarrow{BC}\right)\)
\(\Rightarrow3\overrightarrow{MB}=-2\overrightarrow{BC}\Rightarrow\overrightarrow{BM}=\frac{2}{3}\overrightarrow{BC}=\frac{2}{3}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)=-\frac{2}{3}\overrightarrow{AB}+\frac{2}{3}\overrightarrow{AC}\)
\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}=\overrightarrow{AB}-\frac{2}{3}\overrightarrow{AB}+\frac{2}{3}\overrightarrow{AC}=\frac{1}{3}\overrightarrow{AB}+\frac{2}{3}\overrightarrow{AC}\)
\(\Rightarrow\left\{{}\begin{matrix}m=\frac{1}{3}\\n=\frac{2}{3}\end{matrix}\right.\) \(\Rightarrow mn=\frac{2}{9}\)
Lời giải:
Lấy điểm $N$ trên $AB$ sao cho $MN\parallel AC$
Ta có:
\(\overrightarrow{AM}=\overrightarrow{AN}+\overrightarrow{NM}=\frac{AN}{AB}.\overrightarrow{AB}+\frac{NM}{AC}.\overrightarrow{AC}\)
Mà:
\(\frac{AN}{AB}=\frac{MC}{BC}; \frac{NM}{AC}=\frac{MB}{BC}\) theo định lý Ta-let với $MN\parallel AC$
\(\Rightarrow \overrightarrow{AM}=\frac{MC}{BC}\overrightarrow{AB}+\frac{MB}{BC}\overrightarrow{AC}\)
Ta có đpcm.
\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}\)
\(=\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{BC}\)
\(=\overrightarrow{AB}+\dfrac{2}{3}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)\)
\(=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\)