Cho hai véc tơ $\overrightarrow a $ và $\overrightarrow b $ thỏa mãn các điều kiện $\left| {\overrightarrow a } \right| = \dfrac{1}{2}\left| {\overrightarrow b } \right| = 1$,$\left| {\overrightarrow a  - 2\overrightarrow b } \right| = \sqrt {15} $. Đặt $\overrightarrow u  = \overrightarrow a  + \overrightarrow b $ và $\overrightarrow v  = 2k\overrightarrow a  - \overrightarrow b $, $k \in \mathbb{R}$. Tìm tất cả các giá trị của $k$ sao cho $\left( {\overrightarrow u ,\overrightarrow v } \right) = 60^\circ $

$\left| {\overrightarrow a  - 2\overrightarrow b } \right| = \sqrt {15}  \Leftrightarrow {\left| {\overrightarrow a } \right|^2} + 4{\left| {\overrightarrow b } \right|^2} - 4\overrightarrow a \overrightarrow b  = 15 \Leftrightarrow 2\overrightarrow a \overrightarrow b  = 1$.

$\overrightarrow u \overrightarrow v  = \left( {\overrightarrow a  + \overrightarrow b } \right)\left( {2k\overrightarrow a  - \overrightarrow b } \right) = 2k{\left| {\overrightarrow a } \right|^2} - {\left| {\overrightarrow b } \right|^2} + \left( {2k - 1} \right)\overrightarrow a \overrightarrow b  = 2k - 4 + \dfrac{{2k - 1}}{2}$.

${\left( {\left| {\overrightarrow u } \right|\left| {\overrightarrow v } \right|} \right)^2} = {\left( {\left| {\left( {\overrightarrow a  + \overrightarrow b } \right)} \right|\left| {\left( {2k\overrightarrow a  - \overrightarrow b } \right)} \right|} \right)^2}$$ = \left( {{{\left| {\overrightarrow a } \right|}^2} + {{\left| {\overrightarrow b } \right|}^2} + 2\overrightarrow a \overrightarrow b } \right)\left( {4{k^2}{{\left| {\overrightarrow a } \right|}^2} + {{\left| {\overrightarrow b } \right|}^2} - 4k\overrightarrow a \overrightarrow b } \right)$$ = \left( {5 + 2\overrightarrow a \overrightarrow b } \right)\left( {4{k^2} + 4 - 4k\overrightarrow a \overrightarrow b } \right)$$ = 6\left( {4{k^2} + 4 - 2k} \right)$$ \Rightarrow \left| {\overrightarrow u } \right|\left| {\overrightarrow v } \right| = \sqrt {6\left( {4{k^2} + 4 - 2k} \right)} $.

$\left( {\overrightarrow u ,\overrightarrow v } \right) = 60^\circ $$ \Rightarrow \cos \left( {60^\circ } \right) = \dfrac{{\overrightarrow u \overrightarrow v }}{{\left| {\overrightarrow u } \right|\left| {\overrightarrow v } \right|}}$$ \Leftrightarrow \dfrac{1}{2} = \dfrac{{2k - 4 + \dfrac{{2k - 1}}{2}}}{{\sqrt {6\left( {4{k^2} + 4 - 2k} \right)} }}$$ \Leftrightarrow \sqrt {6\left( {4{k^2} + 4 - 2k} \right)}  = 6k - 9$

$ \Leftrightarrow \sqrt {6\left( {4{k^2} + 4 - 2k} \right)}  = 6k - 9$$ \Leftrightarrow \left\{ \begin{array}{l}k \ge \dfrac{3}{2}\\\sqrt {6\left( {4{k^2} + 2 - k} \right)}  = 6k - 9\end{array} \right.$$ \Leftrightarrow \left\{ \begin{array}{l}k \ge \dfrac{3}{2}\\12{k^2} - 96k + 57 = 0\end{array} \right.$$ \Leftrightarrow \left\{ \begin{array}{l}k \ge \dfrac{3}{2}\\k = 4 \pm \dfrac{{3\sqrt 5 }}{2}\end{array} \right.$$ \Leftrightarrow k = 4 + \dfrac{{3\sqrt 5 }}{2}$.