Cho hàm số \(y = f\left( x \right) = {2^{2018}}{x^3} + {3.

Đồ thị hàm số \(y = f\left( x \right) = {2^{2018}}{x^3} + {3.2^{2018}}{x^2} - 2018\) cắt trục hoành tại 3 điểm phân biệt có hoành độ \({x_1},{x_2},{x_3}\) 

=>  Đặt \(f\left( x \right) = {2^{2018}}\left( {x - {x_1}} \right)\left( {x - {x_2}} \right)\left( {x - {x_3}} \right)\) 

\(\begin{array}{l}
f'\left( x \right) = {2^{2018}}\left[ {\left( {x - {x_2}} \right)\left( {x - {x_3}} \right) + \left( {x - {x_1}} \right)\left( {x - {x_3}} \right) + \left( {x - {x_1}} \right)\left( {x - {x_2}} \right)} \right]\\
 \Rightarrow \left\{ \begin{array}{l}
f'\left( {{x_1}} \right) = {2^{2018}}\left( {x - {x_2}} \right)\left( {x - {x_3}} \right)\\
f'\left( {{x_2}} \right) = {2^{2018}}\left( {x - {x_1}} \right)\left( {x - {x_3}} \right)\\
f'\left( {{x_3}} \right) = {2^{2018}}\left( {x - {x_1}} \right)\left( {x - {x_2}} \right)
\end{array} \right.\\
P = \frac{1}{{f'\left( {{x_1}} \right)}} + \frac{1}{{f'\left( {{x_2}} \right)}} + \frac{1}{{f'\left( {{x_3}} \right)}} = \frac{1}{{{2^{2018}}}}\left( {\frac{1}{{\left( {{x_1} - {x_2}} \right)\left( {{x_1} - {x_3}} \right)}} + \frac{1}{{\left( {{x_2} - {x_1}} \right)\left( {{x_2} - {x_3}} \right)}} + \frac{1}{{\left( {{x_3} - {x_1}} \right)\left( {{x_3} - {x_2}} \right)}}} \right)\\
 = \frac{1}{{{2^{2018}}}}.\frac{{ - \left( {{x_2} - {x_3}} \right) - \left( {{x_3} - {x_1}} \right) - \left( {{x_1} - {x_2}} \right)}}{{\left( {x{}_1 - {x_2}} \right)\left( {{x_2} - {x_3}} \right)\left( {{x_3} - {x_1}} \right)}} = \frac{1}{{{2^{2018}}}}.\frac{0}{{\left( {x{}_1 - {x_2}} \right)\left( {{x_2} - {x_3}} \right)\left( {{x_3} - {x_1}} \right)}} = 0.
\end{array}\)

Vậy P = 0.