Giải thích các bước giải:
a. Ta có:
`10A=(10^1991+10)/(10^1991+1)=1+9/(10^1991+1)`
`10B=(10^1992+10)/(10^1992+1)=1+9/(10^1992+1)`
Vì `1+9/(10^1991+1)>1+9/(10^1992+1)` nên `10A>10B⇔A>B`
Vậy `A>B`
b. Ta có:
`1/2019C=(2019^20+1)/(2019^20+2019)=1-2018/(2019^20+2019)`
`1/2019D=(2019^19+1)/(2019^19+2019)=1-2018/(2019^19+2019)`
Vì `2018/(2019^20+2019)<2018/(2019^19+2019)`
nên `1-2018/(2019^20+2019)>1-2018/(2019^19+2019)`
`⇒1/2019C>1/2019D⇔C>D`
Vậy `C>D`