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Bài giải
\(\overrightarrow{AB}+\overrightarrow{CD}+\overrightarrow{EA}=\overrightarrow{CB}+\overrightarrow{ED}\)
\(\leftrightarrow\text{ }\overrightarrow{AB}-\overrightarrow{CB}+\overrightarrow{CD}-\overrightarrow{ED}+\overrightarrow{EA}=0\)
\(\leftrightarrow\text{ }\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{DC}+\overrightarrow{DE}+\overrightarrow{EA}=0\)
\(\leftrightarrow\text{ }\overrightarrow{AC}+\overrightarrow{CE}+\overrightarrow{EA}=0\)
\(\leftrightarrow\text{ }\overrightarrow{AE}+\overrightarrow{EA}=0\) ( luôn đúng )
\(\Rightarrow\text{ ĐPCM}\)
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.
\(\overrightarrow{w}+\overrightarrow{a}=\overrightarrow{b}-\overrightarrow{c}\Rightarrow\overrightarrow{w}=\overrightarrow{b}-\overrightarrow{c}-\overrightarrow{a}\)
\(\Rightarrow\overrightarrow{w}=\left(3;4\right)-\left(7;2\right)-\left(2;1\right)=\left(-6;1\right)\)
\(m\overrightarrow{a}=m\left(-1;-2\right)=\left(-m;-2m\right)\)
\(n\overrightarrow{b}=n\left(1;-3\right)=\left(n;-3n\right)\)
\(\Rightarrow m\overrightarrow{a}+n\overrightarrow{b}=\left(-m+n;-2m-3n\right)\)
\(\Rightarrow\left\{{}\begin{matrix}-m+n=2\\-2m-3n=-4\end{matrix}\right.\) \(\Rightarrow m-n=-2\) (đảo dấu pt đầu là ra, ko cần giải hẳn ra m; n)
Đặt \(\overrightarrow{b}=x\cdot\overrightarrow{a}+y\cdot\overrightarrow{c}\)
mà \(\overrightarrow{b}=\left(-1;-1\right);\overrightarrow{a}=\left(4;-2\right);\overrightarrow{c}=\left(2;5\right)\)
nên \(\left\{{}\begin{matrix}4x+2y=-1\\-2x+5y=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x+2y=-1\\-4x+10y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}12y=-3\\4x+2y=-1\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-\dfrac{1}{4}\\4x=-1-2y=-1-2\cdot\dfrac{-1}{4}=-1+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=-\dfrac{1}{8}\\y=-\dfrac{1}{4}\end{matrix}\right.\)
Vậy: \(\overrightarrow{b}=\dfrac{-1}{8}\cdot\overrightarrow{a}+\dfrac{-1}{4}\cdot\overrightarrow{c}\)
a/ \(\overrightarrow{u}=2\left(2;3\right)+3\left(4;1\right)-\left(1;1\right)\)
\(\Leftrightarrow\overrightarrow{u}=\left(4+12-1;6+3-1\right)=\left(15;8\right)\)
b/ \(\left(1;1\right)=x\left(2;3\right)+y\left(4;1\right)\)
\(\Rightarrow\left\{{}\begin{matrix}2x+4y=1\\3x+y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{3}{10}\\y=\frac{1}{10}\end{matrix}\right.\Rightarrow\overrightarrow{c}=\frac{3}{10}\overrightarrow{a}+\frac{1}{10}\overrightarrow{b}\)