Gọi \(M\left(m;0\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{MA}=\left(1-m;1\right)\\\overrightarrow{MB}=\left(-2-m;4\right)\end{matrix}\right.\)

\(\Rightarrow\overrightarrow{MA}-2\overrightarrow{MB}=\left(m+5;-7\right)\)

\(\Rightarrow\left|\overrightarrow{MA}-2\overrightarrow{MB}\right|=\sqrt{\left(m+5\right)^2+49}\ge7\)

Dấu "=" xảy ra khi \(m+5=0\Leftrightarrow m=-5\) hay \(M\left(-5;0\right)\)

Cảm ơn rất nhìu ạ

Đáp án B

\(\overrightarrow{u}=\frac12\cdot\overrightarrow{i}-5\cdot\overrightarrow{j}\)

=>\(\overrightarrow{u}=\left(\frac12;-5\right)\)

=>\(\left|\overrightarrow{u}\right|=\sqrt{\left(\frac12\right)^2+\left(-5\right)^2}=\sqrt{25+\frac14}=\sqrt{25,25}\)

\(\overrightarrow{v}=k\cdot\overrightarrow{i}-4\cdot\overrightarrow{j}\)

=>\(\overrightarrow{v}=\left(k;-4\right)\)

=>\(\left|\overrightarrow{v}\right|=\sqrt{k^2+\left(-4\right)^2}=\sqrt{k^2+16}\)

Ta có: \(\left|\overrightarrow{u}\right|=\left|\overrightarrow{v}\right|\)

=>\(k^2+16=25,25\)

=>\(k^2=9,25=\frac{37}{4}\)

=>\(k=\pm\frac{\sqrt{37}}{2}\)

\(\overrightarrow{AB}=\overrightarrow{AC}+\overrightarrow{CB}=\overrightarrow{AC}+\overrightarrow{CD}+\overrightarrow{DB}=\overrightarrow{AC}-\overrightarrow{AB}-\overrightarrow{BD}\)

\(\Rightarrow2\overrightarrow{AB}=\overrightarrow{AC}-\overrightarrow{BD}\Rightarrow\overrightarrow{AB}=\frac{1}{2}\left(\overrightarrow{AC}-\overrightarrow{BD}\right)=\left(5;-\frac{7}{2}\right)\)

A,(2;1) B(-2;-1) C(-5;4) D (5;-4)

Áp dụng quy tắc hình bình hành ta có:

\(\left\{{}\begin{matrix}\overrightarrow{AB}+\overrightarrow{AD}=\widehat{AC}\\\overrightarrow{AD}-\overrightarrow{AB}=\overrightarrow{BD}\end{matrix}\right.\)

Từ hệ trên suy ra:
\(\overrightarrow{2AB}=\left(\overrightarrow{AB}+\overrightarrow{AD}\right)-\left(\overrightarrow{AD}-\overrightarrow{AB}\right)=\overrightarrow{AC}-\overrightarrow{BD}\)

\(\Leftrightarrow\overrightarrow{AB}=\frac{1}{2}\left(\overrightarrow{AC}-\overrightarrow{BD}\right)=\frac{1}{2}\left[7-\left(-3\right);-3-4\right]=\left(5;\frac{-7}{2}\right)\)