Theo tính chất trọng tâm ta có: \(\overrightarrow{AG}=\dfrac{2}{3}\overrightarrow{AM}\)

Mặt khác AM là trung tuyến nên: \(\overrightarrow{AM}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\)

\(\Rightarrow\overrightarrow{AG}=\dfrac{2}{3}\left(\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AC}\right)\Rightarrow3\overrightarrow{AG}=\overrightarrow{AB}+\overrightarrow{AC}\) (1)

K là trung điểm AB, N là trung điểm AC nên: \(\left\{{}\begin{matrix}\overrightarrow{AK}=\dfrac{1}{2}\overrightarrow{AB}\\\overrightarrow{AN}=\dfrac{1}{2}\overrightarrow{AC}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\overrightarrow{AB}=2\overrightarrow{AK}\\\overrightarrow{AC}=2\overrightarrow{AN}\end{matrix}\right.\) (2)

(1);(2) \(\Rightarrow3\overrightarrow{AG}=2\left(\overrightarrow{AK}+\overrightarrow{AN}\right)\)

Chọn A

Ta có 

Do đó 

Đáp án C

A M → = 2 A B → − 3 A C → D N → = D B → + x D C → = A B → − A D → + x A C → − A D → = A B → + x A C → − ( x + 1 ) A D → M N → = A N → − A M → = A D → + D N → − A M → = − A B → + ( x + 3 ) A C → − x A D → B C → = A C → − A B →

Để 3 vectơ A D → , B C → , M N →  đồng phẳng ⇔ ∃ m , n ∈ R  sao cho :

A M → = 2 A B → − 3 A C → D N → = D B → + x D C → = A B → − A D → + x A C → − A D → = A B → + x A C → − ( x + 1 ) A D → M N → = A N → − A M → = A D → + D N → − A M → = − A B → + ( x + 3 ) A C → − x A D → B C → = A C → − A B → M N → = m . A D → + n B C → ⇔ − A B → + ( x + 3 ) A C → − x A D → = m A D → + n ( A C → − A B → ) ⇔ n − 1 = 0 x + 3 − n = 0 x + m = 0 ⇔ n = 1 x = − 2 m = 2