Ta có: `2^2018/(2^2018+3^2019+5^2019)+3^2019/(3^2019 + 5^2020 +2^2018)+ 5^2020/(5^2020+2^2019+3^2019)`
`=(2^2018+3^2019+5^2020)/(2^2018+3^2019+5^2020)`
`=1`
Lại có: `1/(1.2)+1/(2.3)+1/(3.4)+...+(1/2019.2020)`
`=1-1/2020 <1`
Mà: `2^2018/(2^2018+3^2019)+3^2019/(3^2019+5^2020)+5^2020/(5^2020+2^2019) >2^2018/(2^2018+3^2019+5^2019)+3^2019/(3^2019 + 5^2020 +2^2018)+ 5^2020/(5^2020+2^2019+3^2019)`
`=>2^2018/(2^2018+3^2019)+3^2019/(3^2019+5^2020)+5^2020/(5^2020+2^2019)>1(2)`
Và: `1/(1.2)+1/(3.4)+1/(5.6)+...+1/(2019.2020) <1/(1.2)+1/(2.3)+1/(3.4)+...+(1/2019.2020)`
`=>1/(1.2)+1/(3.4)+1/(5.6)+...+1/(2019.2020) <1(2)`
Từ: `(1)+(2)=>A>B`