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Lời giải:
Gọi \(\overrightarrow{d}=(x,y)\). Theo bài ra ta có:
\(\left\{\begin{matrix} \overrightarrow{a}.\overrightarrow{d}=4\\ \overrightarrow{b}.\overrightarrow{d}=-2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} -2x+3y=4\\ 4x+y=-2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=\frac{-5}{7}\\ y=\frac{6}{7}\end{matrix}\right.\)
Vậy.......
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 =))
a.
\(\overrightarrow{u}=2\left(2;1\right)-\left(3;4\right)=\left(1;-2\right)\)
\(\overrightarrow{v}=3\left(3;4\right)-2\left(7;2\right)=\left(-5;8\right)\)
\(\overrightarrow{w}=5\left(7;2\right)+\left(2;1\right)=\left(37;11\right)\)
b.
\(\overrightarrow{x}=2\left(2;1\right)+\left(3;4\right)-\left(7;2\right)=\left(0;4\right)\)
\(\overrightarrow{z}=2\left(2;1\right)-3\left(3;4\right)+\left(7;2\right)=\left(2;-8\right)\)
c.
\(\overrightarrow{w}+\overrightarrow{a}=\overrightarrow{b}-\overrightarrow{c}\Rightarrow\overrightarrow{w}=\overrightarrow{b}-\overrightarrow{c}-\overrightarrow{a}\)
\(\Rightarrow\overrightarrow{w}=\left(3;4\right)-\left(7;2\right)-\left(2;1\right)=\left(-6;1\right)\)
a: 2 vecto IA+vecto IB=vecto 0
=>2 vecto IA=-vecto IB
=>I nằm giữa A và B và IA=2IB
=>vecto AI=2/3*vecto AB
b: 2/5vecto MA+3/5vecto MB
=2/5vecto MI+2/5vecto IA+3/5vecto MI+3/5vecto IB
=vetco MI+1/5(2 vecto IA+3 vecto IB)
=vecto MI