Phương pháp giải:

Sử dụng công thức khai triển của nhị thức : \({\left( {a + b} \right)^n} = \sum\limits_{k = 0}^n {C_n^k{a^{n - k}}{b^k}.} \)

Lời giải chi tiết:

 \(\begin{array}{l}{\left( {\sqrt 3  + x} \right)^{2019}} = \sum\limits_{k = 0}^{2019} {C_{2019}^k{{\left( {\sqrt 3 } \right)}^k}{x^{2019 - k}}} \\ = C_{2019}^0{\left( {\sqrt 3 } \right)^{2019}} + C_{2019}^1{\left( {\sqrt 3 } \right)^{2018}}x + C_{2019}^2{\left( {\sqrt 3 } \right)^{2017}}{x^2} + ...... + C_{2019}^{2018}.\sqrt 3 {x^{2018}} + C_{2019}^{2019}{x^{2019}}\\ = {a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3} + ...... + {a_{2019}}{x^{2019}}.\end{array}\)

Ta có: \({i^m} = \left\{ \begin{array}{l}1\,\,\,\,khi\,\,\,m = 4l\\i\,\,\,khi\,\,\,m = 4l + 1\\- 1\,\,\,\,khi\,\,\,m = 4l + 2\\- i\,\,\,khi\,\,\,m = 4l + 3\end{array} \right.\,\,\left( {l \in Z } \right)\)

Chọn \(x = i\) ta có:

\(\begin{array}{l}
{\left( {\sqrt 3 + i} \right)^{2019}} = \sum\limits_{k = 0}^{2019} {C_{2019}^k{{\left( {\sqrt 3 } \right)}^k}} {i^{2019 - k}}\,\,\,\,\left( {{i^2} = - 1} \right)\\
= C_{2019}^0{\left( {\sqrt 3 } \right)^{2019}} + C_{2019}^1{\left( {\sqrt 3 } \right)^{2018}}i + C_{2019}^2{\left( {\sqrt 3 } \right)^{2017}}{i^2} + ...... + C_{2019}^{2018}.\sqrt 3 {i^{2018}} + C_{2019}^{2019}{i^{2019}}\\
= {a_0} + {a_1}i + {a_2}{i^2} + {a_3}{i^3} + ...... + {a_{2018}}{i^{2018}} + {a_{2019}}{i^{2019}}\\
= {a_0} + {a_1}i - {a_2} - {a_3}i + ...... - {a_{2018}} - {a_{2019}}i.
\end{array}\)

Chọn \(x =  - i\) ta có:

\(\begin{array}{l}
{\left( {\sqrt 3 - i} \right)^{2019}} = \sum\limits_{k = 0}^{2019} {C_{2019}^k{{\left( {\sqrt 3 } \right)}^k}{{\left( { - i} \right)}^{2019 - k}}} \\
= C_{2019}^0{\left( {\sqrt 3 } \right)^{2019}} - C_{2019}^1{\left( {\sqrt 3 } \right)^{2018}}i + C_{2019}^2{\left( {\sqrt 3 } \right)^{2017}}{i^2} - ...... + C_{2019}^{2018}.\sqrt 3 {i^{2018}} - C_{2019}^{2019}{i^{2019}}\\
= {a_0} - {a_1}i + {a_2}{i^2} - {a_3}{i^3} + ...... + {a_{2018}}{i^{2018}} - {a_{2019}}{i^{2019}}\\
= {a_0} - {a_1}i - {a_2} + {a_3}i + ......... - {a_{2018}} + {a_{2019}}i\\
\Rightarrow {\left( {\sqrt 3 + i} \right)^{2019}} + {\left( {\sqrt 3 - i} \right)^{2019}} = 2\left( {{a_0} - {a_2} + {a_4} - {a_6} + ..... + {a_{2016}} - {a_{2018}}} \right)\\
\Leftrightarrow 2S = {\left[ {{{\left( {\sqrt 3 + i} \right)}^3}} \right]^{673}} + {\left[ {{{\left( {\sqrt 3 - i} \right)}^3}} \right]^{673}} = {\left( {8i} \right)^{673}} + {\left( { - 8i} \right)^{673}} = 0\\
\Leftrightarrow 2S = {8^{673}}.{i^{673}} - {8^{673}}.{i^{673}} = 0 \Leftrightarrow S = 0.
\end{array}\)

Chọn B.