Tham khảo
Xét `A=\frac{4-7^{2020}}{7^{2020}}+\frac{5+7^{2021}}{7^{2021}}`
`⇒A=\frac{4}{7^{2020}}-1+\frac{5}{7^{2021}}+1`
`⇒A=\frac{4}{7^{2020}}+\frac{5}{7^{2021}}`
`⇒A=\frac{28}{7^{2021}}+\frac{5}{7^{2021}}`
`⇒A=\frac{33}{7^{2021}}(1)`
Có `B=\frac{1}{7^{2019}}=\frac{7^2}{7^{2021}}=\frac{49}{7^{2021}}(2)`
Từ `(1)(2)⇒A<B(\frac{33}{7^{2021}}<\frac{49}{7^{2021}})`
`\text{©CBT}`
