Đáp án đúng: C
Phương pháp giải:
Áp dụng: \(\left\{ \begin{array}{l}{x_1} + {x_2} + {x_3} = a\\{x_1}{x_2} + {x_2}{x_3} + {x_3}{x_1} = b\\{x_1}{x_2}{x_3} = c\end{array} \right.\) là nghiệm của phương trình bậc ba \({x^3} - a{x^2} + bx - c = 0\).Giải chi tiết:Ta có: \(\left\{ \begin{array}{l}{x_1} + {x_2} + {x_3} = a\\{x_1}{x_2} + {x_2}{x_3} + {x_3}{x_1} = b\\{x_1}{x_2}{x_3} = c\end{array} \right.\) là nghiệm của phương trình bậc ba \({x^3} - a{x^2} + bx - c = 0\).
\(\begin{array}{l}{x_1} + {x_2} + {x_3} = \cos \dfrac{\pi }{7} + \cos \dfrac{{3\pi }}{7} + \cos \dfrac{{5\pi }}{7}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{2\sin \dfrac{\pi }{7}\left( {\cos \dfrac{\pi }{7} + \cos \dfrac{{3\pi }}{7} + \cos \dfrac{{5\pi }}{7}} \right)}}{{2\sin \dfrac{\pi }{7}}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{2\sin \dfrac{\pi }{7}\cos \dfrac{\pi }{7} + 2\sin \dfrac{\pi }{7}\cos \dfrac{{3\pi }}{7} + 2\sin \dfrac{\pi }{7}\cos \dfrac{{5\pi }}{7}}}{{2\sin \dfrac{\pi }{7}}} = \dfrac{{\sin \dfrac{{6\pi }}{7}}}{{2\sin \dfrac{\pi }{7}}} = \dfrac{1}{2}\end{array}\)
\(\begin{array}{l}{x_1}{x_2} + {x_2}{x_3} + {x_3}{x_1} = \cos \dfrac{\pi }{7} \cdot \cos \dfrac{{3\pi }}{7} + \cos \dfrac{{3\pi }}{7} \cdot \cos \dfrac{{5\pi }}{7} + \cos \dfrac{{5\pi }}{7} \cdot \cos \dfrac{\pi }{7}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = c{\rm{os}}\dfrac{{2\pi }}{7} + c{\rm{os}}\dfrac{{4\pi }}{7} + c{\rm{os}}\dfrac{{6\pi }}{7} = - \dfrac{1}{2}\end{array}\)
\(\begin{array}{l}{x_1}{x_2}{x_3} = \cos \dfrac{\pi }{7} \cdot \cos \dfrac{{3\pi }}{7} \cdot \cos \dfrac{{5\pi }}{7} = - \dfrac{1}{8}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{1}{2}\left( {\cos \dfrac{{8\pi }}{7} + \cos \dfrac{{2\pi }}{7}} \right)\cos \dfrac{\pi }{7}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{1}{4}\left( {\cos \dfrac{{9\pi }}{7} + \cos \pi + \cos \dfrac{{3\pi }}{7} + \cos \dfrac{\pi }{7}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{1}{4}\left( {\cos \dfrac{{9\pi }}{7} + \cos \dfrac{{3\pi }}{7} + \cos \dfrac{\pi }{7} - 1} \right) = - \dfrac{1}{8}\end{array}\)
\( \Rightarrow \left\{ \begin{array}{l}{x_1} + {x_2} + {x_3} = \dfrac{1}{2}\\{x_1}{x_2} + {x_2}{x_3} + {x_3}{x_1} = \dfrac{{ - 1}}{2}\\{x_1}{x_2}{x_3} = \dfrac{{ - 1}}{8}\end{array} \right.\)
Vậy phương trình cần tìm là \({x^3} - \dfrac{1}{2}{x^2} - \dfrac{1}{2}x + \dfrac{1}{8} = 0\).
Chọn C.
