Question

So sánh A và B

A=\(\dfrac{4-7^{2020}}{7^{2020}}\)+\(\dfrac{5+7^{2021}}{7^{2021}}\)

B=\(\dfrac{1}{7^{2019}}\)

Answer 1

Ta có:

\(A=\dfrac{7\left(4-7^{2020}\right)}{7^{2021}}+\dfrac{5+7^{2021}}{7^{2021}}\)

\(A=\dfrac{28-7^{2021}+5+7^{2021}}{7^{2021}}=\dfrac{33}{7^{2021}}\)

Ta có: \(B=\dfrac{7^2}{7^{2021}}=\dfrac{49}{7^{2021}}\)

=> B>A