Question

C = \(\dfrac{10^{2021}+1}{10^{2020}+1}\) và D = \(\dfrac{10^{2022}+1}{10^{2021}+1}\)

Answer 1

\(C=\dfrac{10^{2021}+10-9}{10^{2020}+1}=10-\dfrac{9}{10^{2020}+1}\)

\(D=\dfrac{10^{2022}+10-9}{10^{2021}+1}=10-\dfrac{9}{10^{2021}+1}\)

mà \(10^{2020}+1< 10^{2021}+1\)

nên \(-\dfrac{9}{10^{2020}+1}< -\dfrac{9}{10^{2021}+1}\)

hay C<D