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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)
a) Ta có:
\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}\)
\(=\overrightarrow{AB}+k\overrightarrow{BC}\)
\(=\overrightarrow{AB}+k\left(\overrightarrow{AC}-\overrightarrow{AB}\right)\)
\(=\left(1-k\right)\overrightarrow{AB}+k\overrightarrow{AC}\)
b) \(\overrightarrow{NP}=\overrightarrow{AP}-\overrightarrow{AN}\)
\(=\dfrac{2}{3}\overrightarrow{AC}-\dfrac{3}{4}\overrightarrow{AB}\)
Để \(AM\perp NP\)
\(\Rightarrow\overrightarrow{AM}.\overrightarrow{NP}=\overrightarrow{0}\)
\(\Rightarrow\left[\left(1-k\right)\overrightarrow{AB}+k\overrightarrow{AC}\right]\left(-\dfrac{3}{4}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\right)=\overrightarrow{0}\)
\(\Leftrightarrow\dfrac{3\left(k-1\right)}{4}AB^2+\dfrac{2k}{3}AC^2+\dfrac{2\left(1-k\right)}{3}\overrightarrow{AB}.\overrightarrow{AC}-\dfrac{3k}{4}\overrightarrow{AB}.\overrightarrow{AC}=\overrightarrow{0}\)
\(\Leftrightarrow\dfrac{3\left(k-1\right)}{4}AB^2+\dfrac{2k}{3}AB^2+\dfrac{1-k}{3}AB^2-\dfrac{3k}{8}AB^2=0\)
\(\Leftrightarrow AB^2\left[\dfrac{3\left(k-1\right)}{4}+\dfrac{2k}{3}+\dfrac{1-k}{3}-\dfrac{3k}{8}\right]=0\)
\(\Leftrightarrow18\left(k-1\right)+16k+8\left(1-k\right)-9k=0\left(AB>0\right)\)
\(\Leftrightarrow17k=10\)
\(\Leftrightarrow k=\dfrac{10}{17}\)
A B C D I K
a)
- \(\overrightarrow{BI}=\frac{1}{2}\left(\overrightarrow{BA}+\overrightarrow{BD}\right)\) (t/c trung điểm)
\(=\frac{1}{2}\left(\overrightarrow{BA}+\frac{1}{2}\overrightarrow{BC}\right)\)
\(=\frac{1}{2}\overrightarrow{BA}+\frac{1}{4}\overrightarrow{BC}\)
- \(\overrightarrow{BK}=\overrightarrow{BA}+\overrightarrow{AK}\)
\(=\overrightarrow{BA}+\frac{1}{3}\overrightarrow{AC}\)
\(=\overrightarrow{BA}+\frac{1}{3}\left(\overrightarrow{BC}-\overrightarrow{BA}\right)\)
\(=\overrightarrow{BA}+\frac{1}{3}\overrightarrow{BC}-\frac{1}{3}\overrightarrow{BA}\)
\(=\frac{2}{3}\overrightarrow{BA}+\frac{1}{3}\overrightarrow{BC}\)
b) Ta có: \(\overrightarrow{BK}=\frac{2}{3}\overrightarrow{BA}+\frac{1}{3}\overrightarrow{BC}=\frac{4}{3}\left(\frac{1}{2}\overrightarrow{BA}+\frac{1}{4}\overrightarrow{BC}\right)=\frac{4}{3}\overrightarrow{BI}\)
=> B,K,I thẳng hàng
c) \(27\overrightarrow{MA}-8\overrightarrow{MB}=2015\overrightarrow{MC}\)
\(\Leftrightarrow27\left(\overrightarrow{MC}+\overrightarrow{CA}\right)-8\left(\overrightarrow{MC}+\overrightarrow{CB}\right)=2015\overrightarrow{MC}\)
\(\Leftrightarrow27\overrightarrow{MC}+27\overrightarrow{CA}-8\overrightarrow{MC}-8\overrightarrow{CB}-2015\overrightarrow{MC}=\overrightarrow{0}\)
\(\Leftrightarrow-1996\overrightarrow{MC}+27\overrightarrow{CA}-8\overrightarrow{CB}=\overrightarrow{0}\)
\(\Leftrightarrow1996\overrightarrow{CM}=8\overrightarrow{CB}-27\overrightarrow{CA}\)
\(\Leftrightarrow\overrightarrow{CM}=\frac{8\overrightarrow{CB}-27\overrightarrow{CA}}{1996}\)
Vậy: Dựng điểm M sao cho \(\overrightarrow{CM}=\frac{8\overrightarrow{CB}-27\overrightarrow{CA}}{1996}\)
Ta có:
\(\overrightarrow{MN}=\overrightarrow{MA}+\overrightarrow{MB}+4\overrightarrow{MC}\)
\(=6\overrightarrow{MI}+\overrightarrow{IA}+\overrightarrow{IB}+4\overrightarrow{IC}\)
\(=6\overrightarrow{MI}+4\overrightarrow{IG}+4\overrightarrow{IC}\)
\(=6\overrightarrow{MI}\)
\(\Rightarrow M,I,N\) thẳng hàng
Đề thiếu ngay câu đầu nên ko thể giải được:
Sao cho \(?=3MB\)
Mk thêm r đó bạn
3.
\(\overrightarrow{CI}=\frac{1}{2}\overrightarrow{CM}+\frac{1}{2}\overrightarrow{CN}=\frac{1}{2}.\frac{3}{4}\overrightarrow{CB}+\frac{1}{2}.\frac{1}{2}\overrightarrow{CA}=\frac{3}{8}\left(\overrightarrow{CA}+\overrightarrow{AB}\right)+\frac{1}{4}\overrightarrow{CA}\)
\(=\frac{5}{8}\overrightarrow{CA}+\frac{3}{8}\overrightarrow{AB}=\frac{3}{8}\overrightarrow{AB}-\frac{5}{8}\overrightarrow{AC}\)
Đặt \(\overrightarrow{CK}=k.\overrightarrow{CI}=\frac{3k}{8}\overrightarrow{AB}-\frac{5k}{8}\overrightarrow{AC}\)
\(\overrightarrow{BK}=\overrightarrow{BC}+\overrightarrow{CK}=\overrightarrow{BA}+\overrightarrow{AC}+\overrightarrow{CK}=-\overrightarrow{AB}+\overrightarrow{AC}+\frac{3k}{8}\overrightarrow{AB}-\frac{5k}{8}\overrightarrow{AC}\)
\(=\frac{3k-8}{8}\overrightarrow{AB}-\frac{5k-8}{8}\overrightarrow{AC}=-2\left(3k-8\right)\left(-\frac{1}{16}\overrightarrow{AB}+\frac{5k-8}{16\left(3k-8\right)}\overrightarrow{AC}\right)\)
Do B;E;K thẳng hàng nên:
\(\frac{5k-8}{16\left(3k-8\right)}=\frac{1}{3}\Rightarrow k=\frac{104}{33}\)
\(\Rightarrow\frac{KI}{KC}=\frac{71}{104}\)
Cách tính toán là như vậy, còn quá trình tính toán đúng hay sai thì bạn tự tính lại
a.
Câu a đề sai hoặc dữ kiện bạn ghi tiếp tục sai.
Gọi P là trung điểm AB thì \(\overrightarrow{IA}+\overrightarrow{IB}=2\overrightarrow{IP}\) theo t/c trung tuyến
\(\overrightarrow{OA}+\overrightarrow{OB}+2\overrightarrow{OM}=\overrightarrow{OI}+\overrightarrow{IA}+\overrightarrow{OI}+\overrightarrow{IB}+2\left(\overrightarrow{OI}+\overrightarrow{IM}\right)\)
\(=4\overrightarrow{OI}+\overrightarrow{IA}+\overrightarrow{IB}+2\overrightarrow{IM}=4\overrightarrow{OI}+2\left(\overrightarrow{IP}+\overrightarrow{IM}\right)\)
Để tổng này bằng \(4\overrightarrow{OI}\) thì \(\overrightarrow{IP}+\overrightarrow{IM}=0\) đồng nghĩa I là trung điểm MP, đồng nghĩa P trùng N, hoàn toàn vô lý
b.
\(CM=3BM\Rightarrow4\overrightarrow{BM}=\overrightarrow{BC}\)
\(4\overrightarrow{AM}=4\overrightarrow{AB}+4\overrightarrow{BM}=4\overrightarrow{AB}+\overrightarrow{BC}=4\overrightarrow{AB}+\overrightarrow{BA}+\overrightarrow{AC}=3\overrightarrow{AB}+\overrightarrow{AC}\)
c.
Từ câu b \(\Rightarrow\overrightarrow{AM}=\frac{3}{4}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\)
\(\overrightarrow{MN}=\overrightarrow{MA}+\overrightarrow{AN}=-\frac{3}{4}\overrightarrow{AB}-\frac{1}{3}\overrightarrow{AC}+\frac{1}{2}\overrightarrow{AC}=-\frac{3}{4}\overrightarrow{AB}+\frac{1}{6}\overrightarrow{AC}\)
\(\overrightarrow{AE}=\frac{5}{4}\overrightarrow{AM}\Rightarrow\overrightarrow{AM}+\overrightarrow{ME}=\frac{5}{4}\overrightarrow{AM}\Rightarrow\overrightarrow{ME}=\frac{1}{4}\overrightarrow{AM}\)
\(\overrightarrow{BE}=\overrightarrow{BM}+\overrightarrow{ME}=\frac{1}{4}\overrightarrow{BC}+\frac{1}{4}\overrightarrow{AM}=\frac{1}{4}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)+\frac{1}{4}\left(\frac{3}{4}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\right)\)
\(\overrightarrow{BE}=-\frac{1}{16}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}\)