Gọi I là trọng tâm tam giác:

\(\Rightarrow\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}=\overrightarrow{0}\)

Kẻ đường cao AH

\(\Rightarrow AI=\dfrac{2}{3}AH=\dfrac{2}{3}.\dfrac{a\sqrt{3}}{2}=\dfrac{a\sqrt{3}}{3}\)

\(\Rightarrow AI^2=\dfrac{a^2}{3}=BI^2=CI^2\)

\(MA^2+MB^2+MC^2=\left(\overrightarrow{MI}+\overrightarrow{IA}\right)^2+\left(\overrightarrow{MI}+\overrightarrow{IB}\right)^2+\left(\overrightarrow{MI}+\overrightarrow{IC}\right)^2\) \(\Leftrightarrow2a^2=3MI^2+2\overrightarrow{MI}\left(\overrightarrow{IA}+\overrightarrow{IB}+\overrightarrow{IC}\right)+IA^2+IB^2+IC^2\)

\(\Leftrightarrow2a^2=3MI^2+3IA^2\)

\(\Leftrightarrow2a^2=3MI^2+\dfrac{3.a^2}{3}\)

\(\Leftrightarrow MI^2=\dfrac{a^2}{3}\)

\(\Leftrightarrow MI=\dfrac{a\sqrt{3}}{3}\)

\(\Rightarrow M\in\) đường tròn tâm I bán kính \(\dfrac{a\sqrt{3}}{3}\)

a.

Do M là trung điểm OB \(\Rightarrow\overrightarrow{OM}=\dfrac{1}{2}\overrightarrow{OB}\)

\(\Rightarrow\overrightarrow{AM}=\overrightarrow{AO}+\overrightarrow{OM}=-\overrightarrow{OA}+\dfrac{1}{2}\overrightarrow{OB}\)

b.

Do N là trung điểm OC \(\Rightarrow\overrightarrow{ON}=\dfrac{1}{2}\overrightarrow{OC}\)

\(\Rightarrow\overrightarrow{BN}=\overrightarrow{BO}+\overrightarrow{ON}=-\overrightarrow{OB}+\dfrac{1}{2}\overrightarrow{OC}\)

\(\overrightarrow{MN}=\overrightarrow{MO}+\overrightarrow{ON}=-\overrightarrow{OM}+\overrightarrow{ON}=-\dfrac{1}{2}\overrightarrow{OB}+\dfrac{1}{2}\overrightarrow{OC}\)

\(\overrightarrow{AB}=\left(-6;-3\right)=-3\left(2;1\right)\Rightarrow\) đường thẳng AB nhận \(\left(2;1\right)\) là 1 vtcp

Phương trình tham số đường thẳng AB có dạng: \(\left\{{}\begin{matrix}x=5+2t\\y=4+t\end{matrix}\right.\)

Do M thuộc AB nên tọa độ M có dạng \(M\left(5+2t;4+t\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{MA}=\left(-2t;-t\right)\\\overrightarrow{MC}=\left(-2-2t;-6-t\right)\end{matrix}\right.\) \(\Rightarrow\overrightarrow{MA}+\overrightarrow{MC}=\left(-2-4t;-6-2t\right)\)

Đặt \(T=\left|\overrightarrow{MA}+\overrightarrow{MC}\right|=\sqrt{\left(-2-4t\right)^2+\left(-6-2t\right)^2}=\sqrt{20\left(t+1\right)^2+20}\ge\sqrt{20}\)

Dấu "=" xảy ra khi \(t+1=0\Rightarrow t=-1\Rightarrow M\left(3;3\right)\)

\(f\left(x;y\right)=3x^2+y^2-2x-xy+y+3\)

\(=\left(x^2-xy+\dfrac{y^2}{4}\right)+\dfrac{1}{2}\left(4x^2-4x+1\right)+\dfrac{1}{3}\left(\dfrac{9}{4}y^2+3y+1\right)+\dfrac{13}{6}\)

\(=\left(x-\dfrac{y}{2}\right)^2+\dfrac{1}{2}\left(2x-1\right)^2+\dfrac{1}{3}\left(\dfrac{3y}{4}+1\right)^2+\dfrac{13}{6}>0;\forall x;y\)

C,D thuộc (P) nên \(C\left(x_1;x_1^2-4x_1-5\right);D\left(x_2;x_2^2-4x_2-5\right)\)

ABCD là hình bình hành

=>\(\overrightarrow{AB}=\overrightarrow{DC}\)

\(\overrightarrow{AB}=\left(3;-4\right);\overrightarrow{DC}=\left(x_1-x_2;x_1^2-4x_1-5-x_2^2+4x_2+5\right)\)

=>\(\left\{{}\begin{matrix}x_1-x_2=3\\x_1^2-x_2^2-4x_1+4x_2=-4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x_1-x_2=3\\\left(x_1-x_2\right)\left(x_1+x_2-4\right)=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1-x_2=3\\x_1+x_2-4=-\dfrac{4}{x_1-x_2}=-\dfrac{4}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x_1-x_2=3\\x_1+x_2=-\dfrac{4}{3}+4=\dfrac{12}{3}-\dfrac{4}{3}=\dfrac{8}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x_1=3+\dfrac{8}{3}=\dfrac{17}{3}\\x_1+x_2=\dfrac{8}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{17}{6}\\x_2=\dfrac{8}{3}-\dfrac{17}{6}=-\dfrac{1}{6}\end{matrix}\right.\)

Khi x=17/6 thì \(y=x^2-4x-5=\left(\dfrac{17}{6}\right)^2-4\cdot\dfrac{17}{6}-5=-\dfrac{299}{36}\)

Khi x=-1/6 thì \(y=\left(-\dfrac{1}{6}\right)^2-4\cdot\dfrac{-1}{6}-5=\dfrac{1}{36}+\dfrac{2}{3}-5=-\dfrac{155}{36}\)

Vậy: \(C\left(\dfrac{17}{6};-\dfrac{299}{36}\right);D\left(-\dfrac{1}{6};-\dfrac{155}{36}\right)\)

Xét ΔMNP có MJ là đường trung tuyến và G là trọng tâm

nên M,G,J thẳng hàng và \(MG=\dfrac{2}{3}MJ\)

Xét ΔMNP có MJ là đường trung tuyến

nên \(\overrightarrow{MJ}=\dfrac{1}{2}\left(\overrightarrow{MN}+\overrightarrow{MP}\right)\)

=>\(\overrightarrow{MG}=\dfrac{2}{3}\cdot\dfrac{1}{2}\left(\overrightarrow{MN}+\overrightarrow{MP}\right)=\dfrac{1}{3}\overrightarrow{MN}+\dfrac{1}{3}\overrightarrow{MP}\)

=>Chọn C