Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có :
\(\overrightarrow{BP}+\overrightarrow{AN}+\overrightarrow{CM}=\overrightarrow{BC}+\overrightarrow{CP}+\overrightarrow{AB}+\overrightarrow{BN}+\overrightarrow{CA}+\overrightarrow{AM}=\overrightarrow{CP}+\overrightarrow{BN}+\overrightarrow{AM}\)\(=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{1}{3}\overrightarrow{BC}+\dfrac{1}{3}\overrightarrow{CA}\)
\(=\dfrac{1}{3}\left(\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}\right)\)
\(=\dfrac{1}{3}\overrightarrow{0}\)
\(=\overrightarrow{0}\)
\(\RightarrowĐPCM\)
a/ \(\overrightarrow{AN}+\overrightarrow{BP}+\overrightarrow{CM}=\frac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)+\frac{1}{2}\left(\overrightarrow{BC}+\overrightarrow{BA}\right)+\frac{1}{2}\left(\overrightarrow{CA}+\overrightarrow{CB}\right)\)
\(=\frac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{BA}\right)+\frac{1}{2}\left(\overrightarrow{AC}+\overrightarrow{CA}\right)+\frac{1}{2}\left(\overrightarrow{BC}+\overrightarrow{CB}\right)=\overrightarrow{0}\)
b/
Do MN là đường trung bình tam giác ABC \(\Rightarrow\overrightarrow{MN}=\frac{1}{2}\overrightarrow{AC}\)
\(\overrightarrow{AN}=\overrightarrow{AM}+\overrightarrow{MN}=\overrightarrow{AM}+\frac{1}{2}\overrightarrow{AC}=\overrightarrow{AM}+\overrightarrow{AP}\)
c/
\(\overrightarrow{AM}+\overrightarrow{BN}+\overrightarrow{CP}=\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{BC}+\frac{1}{2}\overrightarrow{CA}=\frac{1}{2}\overrightarrow{AC}+\frac{1}{2}\overrightarrow{CA}=\overrightarrow{0}\)
\(\overrightarrow{MN}=\overrightarrow{MA}+\overrightarrow{AN}=-\frac{1}{4}\overrightarrow{AB}+\frac{2}{3}\overrightarrow{AC}\)
\(\overrightarrow{NP}=\overrightarrow{NC}+\overrightarrow{CP}=\frac{1}{3}\overrightarrow{AC}+\frac{1}{5}\overrightarrow{BC}=\frac{1}{3}\overrightarrow{AC}+\frac{1}{5}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)\)
\(=\frac{1}{3}\overrightarrow{AC}-\frac{1}{5}\overrightarrow{AB}+\frac{1}{5}\overrightarrow{AC}=-\frac{1}{5}\overrightarrow{AB}+\frac{8}{15}\overrightarrow{AC}=\frac{4}{5}\left(-\frac{1}{4}\overrightarrow{AB}+\frac{2}{3}\overrightarrow{AC}\right)\)
\(\Rightarrow\overrightarrow{NP}=\frac{4}{5}\overrightarrow{MN}\Rightarrow M;N;P\) thẳng hàng
Đok đề cứ thấy sai sai... Sao cho J lại thoả mãn \(\overrightarrow{BC}=\frac{1}{2}\overrightarrow{AC}-\frac{2}{3}\overrightarrow{AB}\) :))
Có \(\overrightarrow{AN}=\overrightarrow{AB}+\overrightarrow{BN}\)
\(\overrightarrow{BP}=\overrightarrow{BC}+\overrightarrow{CP}\)
\(\overrightarrow{CM}=\overrightarrow{CA}+\overrightarrow{AM}\)
Cộng vế vs vế:
\(\overrightarrow{AN}+\overrightarrow{BP}+\overrightarrow{CM}=\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}+\overrightarrow{BN}+\overrightarrow{CP}+\overrightarrow{AM}\)
\(=\overrightarrow{AC}+\overrightarrow{CA}+\frac{1}{3}\left(\overrightarrow{BC}+\overrightarrow{CA}+\overrightarrow{AB}\right)\)
\(=0+\frac{1}{3}\left(\overrightarrow{BA}+\overrightarrow{AB}\right)=0\) (đpcm)
xin slot để làm