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a)\(\overrightarrow{MN}+\overrightarrow{PQ}=\overrightarrow{MP}+\overrightarrow{PN}+\overrightarrow{PM}+\overrightarrow{MQ}=\overrightarrow{MQ}-\overrightarrow{NP}\)
b)\(\overrightarrow{MQ}+\overrightarrow{NP}=\overrightarrow{MF}+\overrightarrow{FQ}+\overrightarrow{NF}+\overrightarrow{FP}=2\overrightarrow{EF}\)
(vì vecto FM+FN=2FE=>-(FM+FN)=-2FE=>MF+NF=2EF)
a: \(\overrightarrow{MN}+\overrightarrow{NP}+\overrightarrow{PQ}\)
\(=\overrightarrow{MP}+\overrightarrow{PQ}\)
\(=\overrightarrow{MQ}\)
1) Ta có:\(\overrightarrow{AB}+\overrightarrow{DE}-\overrightarrow{DB}+\overrightarrow{BC}=\overrightarrow{AE}+\overrightarrow{BC}=\overrightarrow{AC}+\overrightarrow{CE}+\overrightarrow{BE}+\overrightarrow{EC}\)
\(=\overrightarrow{AC}+\overrightarrow{BE}+\overrightarrow{CE}+\overrightarrow{EC}=\overrightarrow{AC}+\overrightarrow{BE}\left(đpcm\right)\)2) a) Ta có: \(\overrightarrow{AD}+\overrightarrow{BE}+\overrightarrow{CF}=\overrightarrow{AE}+\overrightarrow{ED}+\overrightarrow{BF}+\overrightarrow{FE}+\overrightarrow{CD}+\overrightarrow{DF}\)\(=\overrightarrow{AE}+\overrightarrow{BF}+\overrightarrow{CD}+\overrightarrow{ED}+\overrightarrow{DF}+\overrightarrow{FE}\)
\(=\overrightarrow{AE}+\overrightarrow{BF}+\overrightarrow{CD}\left(đpcm\right)\)
b) Ta có: \(\overrightarrow{AB}+\overrightarrow{CD}=\overrightarrow{AD}+\overrightarrow{DB}+\overrightarrow{CB}+\overrightarrow{BD}\)
\(=\overrightarrow{AD}+\overrightarrow{CB}+\overrightarrow{DB}+\overrightarrow{BD}=\overrightarrow{AD}+\overrightarrow{CB}\left(đpcm\right)\)c) \(\overrightarrow{AB}-\overrightarrow{CD}=\overrightarrow{AB}-\overrightarrow{BD}\)
\(\overrightarrow{AB}+\overrightarrow{DC}=\overrightarrow{AB}+\overrightarrow{DB}\)
Ta có: \(\overrightarrow{AB}+\overrightarrow{DC}=\overrightarrow{AB}+\overrightarrow{DB}+\overrightarrow{BC}\) ( đề bài bị lỗi gì à ?? :v ) hay do mình =))
Câu 1:
\(\overrightarrow{AM}=\overrightarrow{AB}+\overrightarrow{BM}\)
\(=\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{BC}\)
\(=\overrightarrow{AB}+\dfrac{2}{3}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)\)
\(=\dfrac{1}{3}\overrightarrow{AB}+\dfrac{2}{3}\overrightarrow{AC}\)
a/ Theo công thức trọng tâm: \(\left\{{}\begin{matrix}x_G=\frac{x_M+x_N+x_P}{3}=\frac{2}{3}\\y_G=\frac{y_M+y_N+y_P}{3}=-1\\\end{matrix}\right.\) \(\Rightarrow G\left(\frac{2}{3};-1\right)\)
b/ Gọi \(Q\left(x;y\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{MN}=\left(0;-2\right)\\\overrightarrow{QP}=\left(-2-x;-3-y\right)\end{matrix}\right.\)
Để MNPQ là hbh \(\Leftrightarrow\overrightarrow{MN}=\overrightarrow{QP}\)
\(\Rightarrow\left\{{}\begin{matrix}-2-x=0\\-3-y=-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\) \(\Rightarrow Q\left(-2;-1\right)\)
c/ \(\overrightarrow{MQ}=\left(-4;-2\right)\Rightarrow\left|\overrightarrow{MQ}\right|=\sqrt{\left(-4\right)^2+\left(-2\right)^2}=2\sqrt{5}\)
Mà \(MQ=NP\) (tính chất hbh) \(\Rightarrow\left|\overrightarrow{NP}\right|=2\sqrt{5}\)
\(\overrightarrow{NP}=\overrightarrow{MQ}=\left(-4;-2\right)\)
d/ I là trung điểm đường chéo MP
\(\Rightarrow\left\{{}\begin{matrix}x_I=\frac{x_M+x_P}{2}=0\\y_I=\frac{y_M+y_P}{2}=-1\end{matrix}\right.\) \(\Rightarrow I\left(0;-1\right)\)
ta có : \(\overrightarrow{MN}+\overrightarrow{NQ}=\overrightarrow{MQ}+\overrightarrow{QN}+\overrightarrow{NQ}=\overrightarrow{MQ}+\overrightarrow{0}=\overrightarrow{MQ}\ne\overrightarrow{MQ}+\overrightarrow{NP}\)
VẬY kết luận đề sai