Tham khảo
Xét `A`
`⇒10A=\frac{10^{2020}+10}{10^{2020}+1}`
`⇒10A=1+\frac{9}{10^{2020}+1}`
Xét `B`
`⇒10B=\frac{10^{2021}+10}{10^{2021}+1}`
`⇒10B=1+\frac{9}{10^{2021}+1}`
Có `\frac{9}{10^{2020}+1}>\frac{9}{10^{2021}+1}`
`⇒1+\frac{9}{10^{2020}+1}>1+\frac{9}{10^{2021}+1}`
Hay `10A>10B`
`⇒A>B`
`\text{©CBT}`