Đáp án:
4) Trọng tâm là gốc O nên:
$\begin{array}{l}
\left\{ \begin{array}{l}
0 = \dfrac{{{x_A} + {x_B} + {x_C}}}{3}\\
0 = \dfrac{{{y_A} + {y_B} + {y_C}}}{3}
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x_C} = - {x_A} - {x_B} = 2 - 3 = - 1\\
{y_C} = - {y_A} - {y_B} = - 2 - 5 = - 7
\end{array} \right.\\
\Rightarrow C\left( { - 1; - 7} \right)\\
\Rightarrow B\\
5)\overrightarrow a + \overrightarrow b = \left( {3 + \left( { - 1} \right); - 4 + 2} \right) = \left( {2; - 2} \right)\\
\Rightarrow A\\
6)\\
\overrightarrow a - \overrightarrow b = \left( { - 1 - 5;2 - \left( { - 7} \right)} \right) = \left( { - 6;9} \right)\\
\Rightarrow C
\end{array}$
