\(T=\overrightarrow{GA}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)+\overrightarrow{GB}.\overrightarrow{CA}+\overrightarrow{GC}.\overrightarrow{AB}\)

\(=\overrightarrow{AB}\left(\overrightarrow{GC}-\overrightarrow{GA}\right)+\overrightarrow{AC}\left(\overrightarrow{GA}-\overrightarrow{GB}\right)\)

\(=\overrightarrow{AB}\left(\overrightarrow{GC}+\overrightarrow{AG}\right)+\overrightarrow{AC}\left(\overrightarrow{GA}+\overrightarrow{BG}\right)\)

\(=\overrightarrow{AB}.\overrightarrow{AC}+\overrightarrow{AC}.\overrightarrow{BA}\)

\(=0\)

Ta đã biết nếu G' là trọng tâm tam giác ABC thì:
\(\overrightarrow{G'A}+\overrightarrow{G'B}+\overrightarrow{G'C}=\overrightarrow{0}\).
Gỉa sử có điểm G thỏa mãn: \(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}\).
Ta sẽ chứng minh \(G\equiv G'\).
Thật vậy:
\(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}\)
\(\Leftrightarrow3\overrightarrow{GG'}+\overrightarrow{G'A}+\overrightarrow{G'B}+\overrightarrow{G'C}=\overrightarrow{0}\)
\(\Leftrightarrow3\overrightarrow{GG'}=\overrightarrow{0}\)
\(\Leftrightarrow\overrightarrow{GG'}=\overrightarrow{0}\).
Vậy \(G\equiv G'\).

Kéo dài đoạn BM , lấy  thuộc BM sao cho MC' = MG

=> ADCG là hình bình hành

=> GB = 2GM = GC'

Ta có : \(\overrightarrow{GA}+\overrightarrow{GB}=\overrightarrow{GC'}=\overrightarrow{CG}\) (quy tắc hình bình hành)

\(\Rightarrow\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{CG}+\overrightarrow{GC}=\overrightarrow{CC}=\overrightarrow{0}\)

Gọi G \(\in\) trung tuyến AE, D đối xứng với E qua G 

 => BGCD là hình bình hành

=> \(\overrightarrow{GB}\) + \(\overrightarrow{GC}\) = \(\overrightarrow{GD}\)  ( quy tắc HBH) và  \(\overrightarrow{GA}\) +\(\overrightarrow{GD}\) = 0

Ta có:

\(\overrightarrow{GA}\) + \(\overrightarrow{GB}\) + \(\overrightarrow{GC}\) = \(\overrightarrow{GA}\) +\(\overrightarrow{GD}\) = \(\overrightarrow{0}\) (đpcm)

D là điểm nào bạn?

1.

\(\Leftrightarrow x^2-3x+1+\dfrac{\sqrt{3}}{3}\sqrt{\left(x^2+x+1\right)\left(x^2-x+1\right)}=0\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+x+1}=a>0\\\sqrt{x^2-x+1}=b>0\end{matrix}\right.\)

\(\Rightarrow2b^2-a^2+\dfrac{\sqrt{3}}{3}ab=0\)

\(\Leftrightarrow\left(\sqrt{3}b-a\right)\left(2b+\sqrt{3}a\right)=0\)

\(\Leftrightarrow a=\sqrt{3}b\)

\(\Leftrightarrow\sqrt{x^2+x+1}=\sqrt{3}.\sqrt{x^2-x+1}\)

\(\Leftrightarrow x^2+x+1=3x^2-3x+3\)

\(\Leftrightarrow2x^2-4x+2=0\)

\(\Leftrightarrow x=1\)

Bài 2:

Đặt \(AB=x>0\) 

\(AG=\dfrac{1}{2}BC=\dfrac{1}{2}\sqrt{a^2+x^2}\)

\(CG=\dfrac{2}{3}\sqrt{\left(\dfrac{AB}{2}\right)^2+AC^2}=\dfrac{2}{3}\sqrt{\dfrac{x^2}{4}+a^2}\)

\(BG=\dfrac{2}{3}\sqrt{AB^2+\left(\dfrac{AC}{2}\right)^2}=\dfrac{2}{3}\sqrt{x^2+\dfrac{a^2}{4}}\)

Ta có:

\(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}\Leftrightarrow\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{AG}\)

\(\Leftrightarrow GB^2+GC^2+2GB.GC.cos\left(\overrightarrow{GB};\overrightarrow{GC}\right)=AG^2\)

\(\Leftrightarrow cos\left(\overrightarrow{GB};\overrightarrow{GC}\right)=\dfrac{AG^2-BG^2-CG^2}{2GB.GC}\)

\(=\dfrac{\dfrac{a^2+x^2}{4}-\left[\dfrac{4}{9}\left(\dfrac{x^2}{4}+a^2\right)+\dfrac{4}{9}\left(\dfrac{a^2}{4}+x^2\right)\right]}{\dfrac{2}{9}\sqrt{\left(a^2+4x^2\right)\left(x^2+4a^2\right)}}\)

\(=-\dfrac{11}{4}.\dfrac{x^2+a^2}{2\sqrt{\left(a^2+4x^2\right)\left(x^2+4a^2\right)}}\le-\dfrac{11}{4}.\dfrac{x^2+a^2}{5\left(x^2+a^2\right)}=-\dfrac{11}{20}\)

Dấu "=" xảy ra khi \(a=x\Leftrightarrow AB=a\)

\(\Rightarrow\)Vậy chọn đáp án C

Gọi M là trung điểm BC \(\Rightarrow\overrightarrow{MB}+\overrightarrow{MC}=\overrightarrow{0}\)

Ta có:

\(\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\overrightarrow{0}\)

\(\Leftrightarrow\overrightarrow{GM}+\overrightarrow{MA}+\overrightarrow{GM}+\overrightarrow{MB}+\overrightarrow{GM}+\overrightarrow{MC}=\overrightarrow{0}\)

\(\Leftrightarrow3\overrightarrow{GM}+\overrightarrow{MA}=\overrightarrow{0}\)

\(\Leftrightarrow\overrightarrow{GM}=\dfrac{1}{3}\overrightarrow{AM}\)

\(\Leftrightarrow\overrightarrow{GA}+\overrightarrow{AM}=\dfrac{1}{3}\overrightarrow{AM}\)

\(\Leftrightarrow\overrightarrow{AG}=\dfrac{2}{3}\overrightarrow{AM}\)

\(\Rightarrow G\) là trọng tâm tam giác ABC