Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
A = \dfrac{{{{2020}^3} + 1}}{{{{2020}^2} - 2019}} = \dfrac{{{{2020}^3} + {1^3}}}{{{{2020}^2} - 2020 + 1}}\\
= \dfrac{{\left( {2020 + 1} \right).\left( {{{2020}^2} - 2020.1 + {1^2}} \right)}}{{{{2020}^2} - 2020 + 1}} = 2020 + 1 = 2021\\
b,\\
B = \dfrac{{{{2020}^3} - 1}}{{{{2020}^2} + 2021}} = \dfrac{{{{2020}^3} - {1^3}}}{{{{2020}^2} + 2020 + 1}}\\
= \dfrac{{\left( {2020 - 1} \right).\left( {{{2020}^2} + 2020.1 + {1^2}} \right)}}{{{{2020}^2} + 2020 + 1}} = 2020 - 1 = 2019
\end{array}\)
