Đáp án:
`A < B`
Giải thích các bước giải:
`A=(2^2020+1)/(2^2022+1)`
`4A=(2^2022+4)/(2^2022+1)=(2^2022+1+3)/(2^2022+1)`
`=1+3/(2^2022+1)`
$\\$
`B=(2^2019+1)/(2^2021+1)`
`4B=(2^2021+4)/(2^2021+1)=(2^2021+1+3)/(2^2021+1)`
`=1+3/(2^2021+1)`
$\\$
Vì `2^2022+1 > 2^2021+1`
`to 3/(2^2022+1) < 3/(2^2021+1)`
`to 1+ 3/(2^2022+1) < 1+ 3/(2^2021+1)`
`to 4A < 4B`
`to A < B`