a/ Để tứ giác ADCB là hbh

\(\Leftrightarrow\overrightarrow{AD}=\overrightarrow{BC}\Leftrightarrow\left(x_D-x_A;y_D-y_A\right)=\left(x_C-x_B;y_C-y_B\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_D-1=4+2\\y_D-2=4-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_D=7\\y_D=0\end{matrix}\right.\Rightarrow D\left(7;0\right)\)

b/ Có phải đề bài là:

\(2\overrightarrow{EA}-4\overrightarrow{EB}+\overrightarrow{EC}=\overrightarrow{0}?\)

\(\Rightarrow2\left(x_A-x_E;y_A-y_E\right)-4\left(x_B-x_E;y_B-y_E\right)+\left(x_C-x_E;y_C-y_E\right)=0\)

\(\Leftrightarrow2\left(1-x_E;2-y_E\right)-4\left(-2-x_E;6-y_E\right)+\left(4-x_E;4-y_E\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}2-2x_E+8+4x_E+4-x_E=0\\4-2y_E-24+4y_E+4-y_E=0\end{matrix}\right.\)

Bạn tự giải nốt

Kết quả bài này là bao nhiêu vậy ạ?

Vì ABCD là hình bình hành

nên vecto AB=vecto DC

=>\(\left\{{}\begin{matrix}x_C-x_D=x_B-x_A\\y_C-y_D=y_B-y_A\end{matrix}\right.\Leftrightarrow D\left(-4;1\right)\)

\(\overrightarrow{EA}=\left(-1-x;-y\right)\)

\(\overrightarrow{EB}=\left(3-x;1-y\right)\)

\(\overrightarrow{EC}=\left(-x;2-y\right)\)

Theo đề, ta có: \(\left\{{}\begin{matrix}-1-x+3\left(3-x\right)-2\left(-x\right)=0\\-y+3\left(1-y\right)-2\left(2-y\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-1-x+9-3x+2x=0\\-y+3-3y-4+2y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2x+8=0\\-2y-1=0\end{matrix}\right.\Leftrightarrow E\left(4;-\dfrac{1}{2}\right)\)

\(Sao ý cuối là sao sao ấy\)

Sai đề rồi!!

Câu 1 trung tuyến gì hả bạn

trung tuyến CM

1/Gọi \(\overline{M}=x\)

Có:\(2\overrightarrow{MA}+5\overrightarrow{MB}\)\(=2\left(-2-x\right)+5\left(5-x\right)\)\(=21-7x=0\)

\(\Leftrightarrow x=3\)

Vậy \(\overline{M}=3\)

Bài 2,3 ý tưởng tương tự.

#Walker

a) \(\overrightarrow{u}=3\overrightarrow{a}+2\overrightarrow{b}-4\overrightarrow{c}=3\left(2;1\right)+2\left(3;-4\right)-4\left(-7;2\right)\)
\(=\left(6;3\right)+\left(6;-8\right)-\left(-28;8\right)\)
\(=\left(6+6+28;3-8-8\right)=\left(40;-13\right)\).
b) \(\overrightarrow{x}+\overrightarrow{a}=\overrightarrow{b}-\overrightarrow{c}\Leftrightarrow\overrightarrow{x}=\overrightarrow{b}-\overrightarrow{c}-\overrightarrow{a}\)
\(\Leftrightarrow\overrightarrow{x}=\left(3;-4\right)-\left(-7;2\right)-\left(2;1\right)\)
\(\Leftrightarrow\overrightarrow{x}=\left(3+7-2;-4-2-1\right)\)
\(\Leftrightarrow\overrightarrow{x}=\left(8;-7\right)\).
c) Có \(\overrightarrow{c}\left(-7;2\right)=k\overrightarrow{a}+h\overrightarrow{b}=k\left(2;1\right)+h\left(3;-4\right)\)
\(=\left(2k+3h;k-4h\right)\).
Từ đó suy ra: \(\left\{{}\begin{matrix}2k+3h=-7\\k-4h=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}k=-2\\h=-1\end{matrix}\right.\).