Trong mặt phẳng tọa độ Oxy, cho các điểm 

D

- Hướng dẫn giải

Ta có: 

\(\left\{ \begin{array}{l}\overrightarrow {AB} = \left( {2 - m;\,\,2 - 2m} \right)\\\overrightarrow {AC} = \left( {2m + 1;\,\, - \dfrac{4}{3}} \right)\end{array} \right.\)

Ba điểm  A, B, C thẳng hàng \(\Leftrightarrow \overrightarrow {AB} = k\overrightarrow {AC} \,\,\,\left( {k \in \mathbb{R},\,\,k \ne 0} \right)\)

\(\begin{array}{l} \Leftrightarrow \left( {2 - m;\,\,2 - 2m} \right) = k\left( {2m + 1; - \dfrac{4}{3}} \right)\\ \Leftrightarrow \left\{ \begin{array}{l}2 - m = k\left( {2m + 1} \right)\\2 - 2m = - \dfrac{4}{3}k\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}k = \dfrac{{3\left( {m - 1} \right)}}{2}\\2 - m = \dfrac{{3\left( {m - 1} \right)}}{2}\left( {2m + 1} \right)\,\,\,\,\left( * \right)\end{array} \right.\\ \Rightarrow \left( * \right) \Leftrightarrow 4 - 2m = 6{m^2} + 3m - 6m - 3\\ \Leftrightarrow 6{m^2} - m - 7 = 0\\ \Leftrightarrow \left( {6m - 7} \right)\left( {m + 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}6m - 7 = 0\\m + 1 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}m = \dfrac{7}{6}\\m = - 1\end{array} \right.\\ \Rightarrow {m_1} + {m_2} = \dfrac{7}{6} - 1 = \dfrac{1}{6}.\end{array}\)