Chọn phương án A.

Ta có:

• \(\left(1+\sqrt{3}\right)^2=4+2\sqrt{3}\)
• \(\left(1+\sqrt{3}\right)\cdot\left(1-\sqrt{3}\right)=-2\)

Vậy đặt \(t=1+\sqrt{3}\) ta được:

• \(4+2\sqrt{3}=t^2\)
• \(1-\sqrt{3}=-\dfrac{2}{t}=-2t^{-1}\)

Khi đó:$$\begin{aligned}
P&=\dfrac{\left(4+2\sqrt{3}\right)^{2018}\cdot\left(1-\sqrt{3}\right)^{2017}}{\left(1+\sqrt{3}\right)^{2019}}\\
&=\dfrac{\left(t^2\right)^{2018}\cdot\left(-2t^{-1}\right)^{2017}}{t^{2019}}\\
&=\dfrac{t^{4036}\cdot(-2)^{2017}t^{-2017}}{t^{2019}}\\
&=(-2)^{2017}t^{4036-2017-2019}\\
&=(-2)^{2017}t^0\\
&=(-2)^{2017}.
\end{aligned}$$