Đặt 2020-x=a
Phương trình trở thành:
\(a^3+\left(a+1\right)^3-\left(2a+1\right)^3=0\)
\(\Leftrightarrow a^3+a^3+3a^2+3a+1-\left(8a^3+12a^2+6a+1\right)=0\)
\(\Leftrightarrow2a^3+3a^2+3a+1-8a^3-12a^2-6a-1=0\)
\(\Leftrightarrow-6a^3-9a^2-3a=0\)
\(\Leftrightarrow-3a\left(2a^2+3a+1\right)=0\)
\(\Leftrightarrow a\left(2a+1\right)\left(a+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\\2a+1=0\\a+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=0\\2a=-1\\a=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=0\\a=-\dfrac{1}{2}\\a=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2020-x=0\\2020-x=-\dfrac{1}{2}\\2020-x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{4041}{2}\\x=2021\end{matrix}\right.\)
Vậy: \(S=\left\{2020;\dfrac{4041}{2};2021\right\}\)
