Đáp án:
\(A = \dfrac{2021}{2}\)
Giải thích các bước giải:
Nhận thấy:
$\dfrac{1}{2022} + \dfrac{2021}{2022} = 1$
$\dfrac{2}{2022} + \dfrac{2020}{2022} = 1$
$\cdots$
Xét: $f(x) + f(1 - x)$
$= \dfrac{x^3}{1-3x + 3x^2} + \dfrac{(1-x)^3}{1 - 3(1-x) + 3(1-x)^2}$
$= \dfrac{x^3}{1-3x + 3x^2} + \dfrac{(1-x)^3}{1 - 3x + 3x^2}$
$= \dfrac{x^3 + (1 - x)^3}{1 - 3x + 3x^2}$
$= \dfrac{(x + 1 - x)[x^2 - x(1-x) + (1-x)^2]}{1 - 3x + 3x^2}$
$= \dfrac{3x^2 -3x + 1}{1 - 3x + 3x^2}$
$= 1$
Ta được:
\(\begin{array}{l}
A = f\left(\dfrac{1}{2022}\right) + f\left(\dfrac{2}{2022}\right) + f\left(\dfrac{3}{2022}\right)+\cdots+ f\left(\dfrac{2020}{2022}\right)+ f\left(\dfrac{2021}{2022}\right)\\
\quad = \left[ f\left(\dfrac{1}{2022}\right) + f\left(\dfrac{2021}{2022}\right)\right] + \left[ f\left(\dfrac{2}{2022}\right) + f\left(\dfrac{2020}{2022}\right)\right] + \cdots +\left[ f\left(\dfrac{1010}{2022}\right)+f\left(\dfrac{1012}{2022}\right)\right] +f\left(\dfrac{1011}{2022}\right)\\
\quad =\underbrace{1+1+\cdots+1}_{\text{1010 số hạng}} + f\left(\dfrac12\right)\\
\quad = 1010+\dfrac12\\\quad = \dfrac{2021}{2}
\end{array}\)
