$\begin{array}{l}1a)\quad A(-2;1)\quad B(4;3)\\ \text{Gọi $I(a;b)$ là trung điểm $AB$}\\ \to \begin{cases}a =\dfrac{-2 + 4}{2} =1\\b = \dfrac{1 + 3}{2} = 2\end{cases}\\ \to I(1;2)\\ b)\quad \text{Ta có:}\quad \overrightarrow{OI} = (1;2)\\ \to \overrightarrow{u} = 2\overrightarrow{OI} \Leftrightarrow \begin{cases}x_u = 2.1 = 2\\y_u = 2.2 = 4\end{cases}\\ \to \overrightarrow{u} = (2;4)\\ 2)\quad Goi\,\,\overrightarrow{u} = (x;y)\\ \quad \overrightarrow{u} = \overrightarrow{BC} + \overrightarrow{AG}\\ \Leftrightarrow \overrightarrow{u} = \overrightarrow{BA} + \overrightarrow{AC} + \dfrac{2}{3}\overrightarrow{AI}\\ \Leftrightarrow \overrightarrow{u} = \overrightarrow{BA} + \overrightarrow{AC} + \dfrac{2}{3}\cdot\dfrac12(\overrightarrow{AB} + \overrightarrow{AC})\\ \Leftrightarrow \overrightarrow{u} =\dfrac{1}{3}\overrightarrow{AB} - \overrightarrow{AB} + \dfrac{1}{3}\overrightarrow{AC} + \overrightarrow{AC}\\ \Leftrightarrow \overrightarrow{u} =-\dfrac{2}{3}\overrightarrow{AB} + \dfrac{4}{3}\overrightarrow{AC}\\ \Leftrightarrow \begin{cases}x = -\dfrac{2}{3}\cdot 6 + \dfrac{4}{3}\cdot(-6)\\y = -\dfrac{2}{3}\cdot(-6) + \dfrac{4}{3}\cdot12 \end{cases}\\ \Leftrightarrow \begin{cases}x = -12\\y = 20\end{cases}\\ Vậy\,\,\overrightarrow{u} = (-12;20) \end{array}$
