gọi M có tọa độ là (x;y) do M thuộc Ox=> tọa ddoooj M là (x;0)

ta có : \(\left|\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}\right|=\left|\left(-2-X;5\right)+\left(3-X;-1\right)+\left(7-X;1\right)\right|\)

=\(\left|\sqrt{\left(-2-X\right)^2+5^2}+\sqrt{\left(3-X\right)^2+1}+\sqrt{\left(7-X\right)^2+1}\right|\)

=> BẠN TÌ gtnn CÁI TRONG LÀ ĐC 

\(d'=T_{\overrightarrow{v}}\left(d\right)\)

Ta có: \(\left\{{}\begin{matrix}x'=x+a\\y'=y+b\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=x'-a=x'-3\\y=y'-b=y'-4\end{matrix}\right.\)

Thay vào pt \(\left(d\right):x+y-6=0\) ta đc:

\(\Rightarrow\left(x'-3\right)+\left(y'-4\right)-6=0\)

\(\Rightarrow x'+y'-13=0\)

Vậy \(\left(d'\right):x+y-13=0\)

\(\left\{{}\begin{matrix}\overrightarrow{MN}\left(x_N-x_M;y_N-y_M\right)=\left(5-x_M;-3-y_M\right)\\\overrightarrow{MP}\left(x_P-x_M;y_P-y_M\right)=\left(1-x_M;-y_M\right)\end{matrix}\right.\)

\(\Rightarrow\overrightarrow{MN}-\overrightarrow{MP}=\left(5-x_M;-3-y_M\right)-\left(1-x_M;-y_M\right)\)

\(=\left(5-x_M-1+x_M;-3-y_M+y_M\right)=\left(4;-3\right)\)

\(cos\left(\overrightarrow{a},\overrightarrow{b}\right)=\dfrac{1\cdot\left(-1\right)+\left(-2\right)\cdot\left(-3\right)}{\sqrt{1^2+2^2}\cdot\sqrt{1^2+3^2}}=\dfrac{5}{\sqrt{5}\cdot\sqrt{10}}=\dfrac{5}{\sqrt{50}}=\dfrac{1}{\sqrt{2}}\)

Gọi tọa độ điểm \(M\) là \(M\left(x;y\right).\)

\(\overrightarrow{MA}=\left(1-x;3-y\right);\overrightarrow{MB}=\left(4-x;-y\right);\overrightarrow{MC}=\left(2-x;-5-y\right).\)

Ta có: \(\overrightarrow{MA}+\overrightarrow{MB}-3\overrightarrow{MC}=\overrightarrow{0}.\)

\(\left\{{}\begin{matrix}1-x+4-x-3\left(2-x\right)=0.\\3-y-y-3\left(-5-y\right)=0.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-2x+5-6+3x=0.\\3-2y+15+3y=0.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0.\\y+18=0.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1.\\y=-18.\end{matrix}\right.\) \(\Rightarrow M\left(1;-18\right).\)