Đáp án + Giải thích các bước giải:

$f(x) = \dfrac{x^3}{1 - 3x + 3x^2}$ 

$= \dfrac{x^3}{x^3 - (x^3 - 3x^2 + 3x - 1)}$
$= \dfrac{x^3}{x^3 - (x - 1)^3}$

$\Rightarrow f(x) + f(1 - x) = \dfrac{x^3}{x^3 - (x - 1)^3} + \dfrac{(1 - x)^3}{(1 - x)^3 + x^3}$

$= \dfrac{x^3}{x^3 + (1 - x)^3} + \dfrac{(1 - x)^3}{x^3 + (1 - x)^3}$

$= \dfrac{x^3 + (1 - x)^3}{x^3 + (1 - x)^3}$

$= 1$

$\Rightarrow P = f\bigg(\dfrac{1}{2024}\bigg) + f\bigg(\dfrac{2}{2024}\bigg) + ... + f\bigg(\dfrac{2023}{2024}\bigg)$

$= f\bigg(\dfrac{1}{2024}\bigg) + f\bigg(\dfrac{2023}{2024}\bigg) + f\bigg(\dfrac{2}{2024}\bigg) + f\bigg(\dfrac{2022}{2024}\bigg) + ... + f\bigg(\dfrac{1011}{2024}\bigg) + f\bigg(\dfrac{1013}{2024}\bigg) + f\bigg(\dfrac{1012}{2024}\bigg)$

$= 1011 + f\bigg(\dfrac{1}{2}\bigg)$

$= 1011 + \dfrac{1}{2}$
$= \dfrac{2023}{2}$