Gọi \(M\left(a;b\right)\)

\(\Rightarrow\overrightarrow{MB}=\left(2-a;3-b\right)\Rightarrow2\overrightarrow{MB}=\left(4-2a;6-2b\right)\)

\(\overrightarrow{MC}=\left(-1-a;-2-b\right)\Rightarrow3\overrightarrow{MC}=\left(-3-3a;-6-3b\right)\)

\(\Rightarrow2\overrightarrow{MB}+3\overrightarrow{MC}=\left(1-5a;-5b\right)=\overrightarrow{0}\)

\(\Rightarrow\left\{{}\begin{matrix}1-5a=0\\-5b=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{1}{5}\\b=0\end{matrix}\right.\) \(\Rightarrow M\left(\frac{1}{5};0\right)\)

M(x;y); A(1;3); B(4;0); C(2;-5)

\(\overrightarrow{MA}=\left(1-x;3-y\right);\overrightarrow{MB}=\left(4-x;0-y\right)=\left(4-x;-y\right)\) ; \(\overrightarrow{MC}=\left(2-x;-5-y\right)\)

\(\overrightarrow{MA}+\overrightarrow{MB}-3\cdot\overrightarrow{MC}=\overrightarrow{0}\)

=>\(\begin{cases}1-x+4-x-3\left(2-x\right)=0\\ 3-y-y-3\left(-5-y\right)=0\end{cases}\Rightarrow\begin{cases}-2x+5-6+3x=0\\ 3-2y+15+3y=0\end{cases}\)

=>\(\begin{cases}x-1=0\\ y+18=0\end{cases}\Rightarrow\begin{cases}x=1\\ y=-18\end{cases}\)

=>M(1;-18)

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

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

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

Chọn A.

M nằm trên trục hoành

=>M(x;0)

M(x;0); B(2;2); C(0;1)

=>\(MB=\sqrt{\left(2-x\right)^2+\left(2-0\right)^2}=\sqrt{\left(x-2\right)^2+4}\)

\(MC=\sqrt{\left(0-x\right)^2+\left(1-0\right)^2}=\sqrt{\left(-x\right)^2+1^2}=\sqrt{x^2+1}\)

MB=2MC

=>\(\sqrt{\left(x-2\right)^2+4}=2\cdot\sqrt{x^2+1}\)

=>\(4\left(x^2+1\right)=\left(x-2\right)^2+4\)

=>\(4x^2+4=x^2-4x+4+4\)

=>\(4x^2+4=x^2-4x+8\)

=>\(3x^2+4x-4=0\)

=>\(3x^2+6x-2x-4=0\)

=>(x+2)(3x-2)=0

=>x=2/3 hoặc x=-2

=>M(2/3;0); M(-2;0)