Đáp án:

Tổng là: $\dfrac{{{{1285}^2}\pi }}{4}$ Giải thích: Ta có: $\eqalign{ & {\sin ^{2018}}x + {\cos ^{2018}}x = 2{\sin ^{2020}}x + 2{\cos ^{2020}}x \cr & \Leftrightarrow {\sin ^{2018}}x\left( {1 - 2{{\sin }^2}x} \right) = {\cos ^{2018}}x\left( {2{{\cos }^2}x - 1} \right) \cr & \Leftrightarrow {\sin ^{2018}}x\cos 2x = {\cos ^{2018}}x\cos 2x \cr & \Leftrightarrow \cos 2x\left( {{{\sin }^{2018}}x - {{\cos }^{2018}}x} \right) = 0 \cr & \Leftrightarrow \left[ \matrix{ \cos 2x = 0 \hfill \cr {\sin ^{2018}}x = {\cos ^{2018}}x \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ \cos 2x = 0 \hfill \cr \sin x = \cos x \hfill \cr \sin x = - \cos x \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ 2x = {\pi \over 2} + k\pi \hfill \cr \tan x = 1 \hfill \cr \tan x = - 1 \hfill \cr} \right. \cr & \Leftrightarrow \left[ \matrix{ x = {\pi \over 4} + {{k\pi } \over 2} \hfill \cr x = {\pi \over 4} + k\pi \hfill \cr x = - {\pi \over 4} + k\pi \hfill \cr} \right. \Leftrightarrow x = {\pi \over 4} + {{k\pi } \over 2}\,\,\left( {k \in Z} \right) \cr} $ \(\begin{array}{l} x = \frac{\pi }{4} + \frac{{k\pi }}{2}\\ x \in \left( {0;2018} \right) \Leftrightarrow 0 < \frac{\pi }{4} + \frac{{k\pi }}{2} < 2018\\ \Rightarrow - \dfrac{1}{2} < k < 1284,2\\ \text{Mà}\,\,k \in Z \Leftrightarrow k \in \left\{ {0;1;2;...;1284} \right\}\\ \Leftrightarrow \text{Tổng} = \dfrac{\pi }{4} + \dfrac{\pi }{4} + \dfrac{\pi }{2} + \dfrac{\pi }{4} + \dfrac{{2\pi }}{2} + ... + \dfrac{\pi }{4} + \dfrac{{1284\pi }}{2}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1285\dfrac{\pi }{4} + \dfrac{\pi }{2}\left( {1 + 2 + 3 + ... + 1284} \right)\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{1285\pi }}{4} + \dfrac{\pi }{2}.\dfrac{{1285.1284}}{2}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{1285\pi }}{4} + \dfrac{{1285\pi }}{4}.1284\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{1285\pi }}{4}\left( {1 + 1284} \right)\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{{{1285}^2}\pi }}{4} \end{array}\)