\(A=\frac{4-7^{2020}}{7^{2020}}+\frac{5+7^{2021}}{7^{2021}}=\frac{4}{7^{2020}}-1+\frac{5}{7^{2021}}+1=\frac{4.7+5}{7^{2021}}=\frac{33}{7^{2021}}\)
\(B=\frac{1}{7^{2019}}=\frac{7^2}{7^{2021}}=\frac{49}{7^{2021}}\)
Có \(\frac{49}{7^{2021}}>\frac{33}{7^{2021}}\Rightarrow A< B\).