Đáp án:
$\\$
`A = (10^{2019} + 1)/(10^{2020} + 1)`
Nhân cả 2 vế với `10` ta được :
`-> 10A = 10 × (10^{2019} + 1)/(10^{2020} + 1)`
`-> 10A = (10^{2020} + 10)/(10^{2020}+1)`
`->10A = (10^{2020} + 1+9)/(10^{2020}+1)`
`-> 10A = (10^{2020}+1)/(10^{2020}+1) + 9/(10^{2020}+1)`
`-> 10A = 1 + 9/(10^{2020}+1)`
`B = (10^{2020}+1)/(10^{2021}+1)`
Nhân cả 2 vế với `10` ta được :
`-> 10B =10 × (10^{2020}+1)/(10^{2021}+1)`
`-> 10B = (10^{2021} + 10)/(10^{2021}+1)`
`-> 10B = (10^{2021}+1+9)/(10^{2021}+1)`
`->10B=(10^{2021}+1)/(10^{2021}+1) + 9/(10^{2021}+1)`
`->10B=1+9/(10^{2021}+1)`
Có : `10^{2020} < 10^{2021}` (Vì `2020 < 2021`)
`-> 10^{2020} +1 < 10^{2021} + 1`
`-> 1/(10^{2020}+1) > 1/(10^{2021} + 1)`
`-> 9/(10^{2020} + 1) > 9/(10^{2021} + 1)`
`-> 1 + 9/(10^{2020} + 1) > 1 + 9/(10^{2021}+1)`
`-> 10A > 10B`
`->A>B`
Vậy `A > B`
