Câu hỏi 39

Ta có: \(\overrightarrow a  = \left( {1;\,\,2;\,\,1} \right),\,\,\overrightarrow b  = \left( { - 2;\,\,m;\,\,3} \right);\,\,\,\angle \left( {\overrightarrow a ;\,\,\overrightarrow b } \right) = {60^0}\)

\(\begin{array}{l} \Rightarrow \cos {60^0} = \dfrac{{1.\left( { - 2} \right) + 2.m + 1.3}}{{\sqrt {1 + {2^2} + 1} .\sqrt {{{\left( { - 2} \right)}^2} + {m^2} + {3^2}} }} = \dfrac{1}{2}\\ \Leftrightarrow \dfrac{{2m + 1}}{{\sqrt {6\left( {13 + {m^2}} \right)} }} = \dfrac{1}{2} \Leftrightarrow 4m + 2 = \sqrt {6\left( {13 + {m^2}} \right)} \\ \Leftrightarrow \left\{ \begin{array}{l}4m + 2 \ge 0\\{\left( {4m + 2} \right)^2} = 6\left( {13 + {m^2}} \right)\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m \ge  - \dfrac{1}{2}\\16{m^2} + 16m + 4 = 78 + 6{m^2}\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}m \ge  - \dfrac{1}{2}\\10{m^2} + 16m - 74 = 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m \ge  - \dfrac{1}{2}\\\left[ \begin{array}{l}m = \dfrac{{ - 4 + \sqrt {201} }}{5}\\m = \dfrac{{ - 4 - \sqrt {201} }}{5}\end{array} \right.\end{array} \right. \Leftrightarrow m = \dfrac{{ - 4 + \sqrt {201} }}{5}.\\ \Rightarrow \left\{ \begin{array}{l}a =  - 4\\b = 201\end{array} \right. \Rightarrow a + b =  - 4 + 201 = 197.\end{array}\)

Chọn  C.