Đáp án:
`A>B`
Giải thích các bước giải:
Ta có:
`A=(10^2019+1)/(10^2020+1)`
`=>10A=(10(10^2019+1))/(10^2020+1)=(10^2020+10)/(10^2020+1)=((10^2020+1)+9)/(10^2020+1)=1+9/(10^2020+1)`
`B=(10^2020+1)/(10^2021+1)`
`=>10B=(10(10^2020+1))/(10^2021+1)=(10^2021+10)/(10^2021+1)=((10^2021+1)+9)/(10^2021+1)=1+9/(10^2021+1)`
Có: `10^2020<10^2021`
`=>10^2020+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.`
