Cho phương trình \({\left( {\dfrac{{\tan \frac{\pi...

D

- Hướng dẫn giải

Phương pháp giải:

\(\tan \left( {a + b} \right) = \dfrac{{\tan \,a + \tan b}}{{1 - \tan \,a\,\tan b}},\,\,\,\,\,\tan \left( {a - b} \right) = \dfrac{{\tan \,a - \tan b}}{{1 + \tan \,a\,\tan b}}\)

Giải chi tiết:

\(\tan \dfrac{\pi }{{12}} = \tan \left( {\dfrac{\pi }{4} - \dfrac{\pi }{6}} \right) = \dfrac{{\tan \,\frac{\pi }{4} - \tan \frac{\pi }{6}}}{{1 + \tan \frac{\pi }{4}\,\tan \frac{\pi }{6}}} \)

\(= \dfrac{{1 - \frac{1}{{\sqrt 3 }}}}{{1 + 1.\frac{1}{{\sqrt 3 }}}} = \dfrac{{\sqrt 3  - 1}}{{\sqrt 3  + 1}} = \dfrac{{{{\left( {\sqrt 3  - 1} \right)}^2}}}{{\left( {\sqrt 3  - 1} \right)\left( {\sqrt 3  + 1} \right)}} = \dfrac{{4 - 2\sqrt 3 }}{2} = 2 - \sqrt 3 \)

Phương trình đã cho tương đương với:

\(\begin{array}{l}{\left( {\dfrac{{2 - \sqrt 3 }}{{1 - \left( {2 - \sqrt 3 } \right)}}} \right)^{\frac{x}{{2017}}}} + \dfrac{{\sqrt[4]{{12}}.\left( {2 - \sqrt 3 } \right)}}{{1 - \left( {2 - \sqrt 3 } \right)}}.{\left( {\dfrac{1}{{1 + \left( {2 - \sqrt 3 } \right)}}} \right)^{\frac{x}{{2017}}}} = 2017.{\left( {\dfrac{1}{{\sqrt[4]{{12}}}}} \right)^{\frac{x}{{2017}}}}\\ \Leftrightarrow {\left( {\dfrac{{\sqrt 3  - 1}}{2}} \right)^{\frac{x}{{2017}}}} + \dfrac{{\sqrt[4]{{12}}.\left( {\sqrt 3  - 1} \right)}}{2}.{\left( {\dfrac{1}{{3 - \sqrt 3 }}} \right)^{\frac{x}{{2017}}}} = 2017.{\left( {\dfrac{1}{{\sqrt[4]{{12}}}}} \right)^{\frac{x}{{2017}}}}\\ \Leftrightarrow {\left( {\dfrac{{\left( {\sqrt 3  - 1} \right)\sqrt[4]{{12}}}}{2}} \right)^{\frac{x}{{2017}}}} + \dfrac{{\sqrt[4]{{12}}.\left( {\sqrt 3  - 1} \right)}}{2}.{\left( {\dfrac{{\sqrt[4]{{12}}}}{{3 - \sqrt 3 }}} \right)^{\frac{x}{{2017}}}} = 2017\end{array}\)

Do \(\,\left( {\dfrac{{\left( {\sqrt 3  - 1} \right)\sqrt[4]{{12}}}}{2}} \right)\left( {\dfrac{{\sqrt[4]{{12}}}}{{3 - \sqrt 3 }}} \right) = \dfrac{{\sqrt {12} }}{{2\sqrt 3 }} = 1\) nên đặt  \({\left( {\dfrac{{\left( {\sqrt 3  - 1} \right)\sqrt[4]{{12}}}}{2}} \right)^{\frac{x}{{2017}}}} = t,\,\,\left( {t > 0} \right) \Rightarrow {\left( {\dfrac{{\sqrt[4]{{12}}}}{{3 - \sqrt 3 }}} \right)^{\frac{x}{{2017}}}} = \dfrac{1}{t}\)

\( \Rightarrow t + \dfrac{{\sqrt[4]{{12}}.\left( {\sqrt 3  - 1} \right)}}{2}.\dfrac{1}{t} = 2017 \Leftrightarrow 2{t^2} - 4034t + \sqrt[4]{{12}}.\left( {\sqrt 3  - 1} \right) = 0\) (1)

Giả sử \({t_1},\,\,{t_2}\) là nghiệm của phương trình (1). Theo Vi ét:  \({t_1}{t_2} = \dfrac{{\sqrt[4]{{12}}.\left( {\sqrt 3  - 1} \right)}}{2}\)

Khi đó:

\(\begin{array}{l}{\left( {\dfrac{{\left( {\sqrt 3  - 1} \right)\sqrt[4]{{12}}}}{2}} \right)^{\frac{{{x_1}}}{{2017}}}}.{\left( {\dfrac{{\left( {\sqrt 3  - 1} \right)\sqrt[4]{{12}}}}{2}} \right)^{\frac{{{x_2}}}{{2017}}}} = \dfrac{{\sqrt[4]{{12}}.\left( {\sqrt 3  - 1} \right)}}{2}\\ \Leftrightarrow {\left( {\dfrac{{\left( {\sqrt 3  - 1} \right)\sqrt[4]{{12}}}}{2}} \right)^{\frac{{{x_1} + {x_2}}}{{2017}}}} = \dfrac{{\sqrt[4]{{12}}.\left( {\sqrt 3  - 1} \right)}}{2}\\ \Leftrightarrow {x_1} + {x_2} = 2017\end{array}\).

Chọn D.