bài 1 : cho E =2019^2019+1/2019^2020+1 cho F = 2019^2020+1/2019^2021+1 so sánh e và f mong mn giúp ạ.help me 07/11/2021 Bởi Clara bài 1 : cho E =2019^2019+1/2019^2020+1 cho F = 2019^2020+1/2019^2021+1 so sánh e và f mong mn giúp ạ.help me
` E = (2019^(2019) +1)/(2019^(2020) +1` ` F = (2019^(2020) +1)/(2019^(2021) +1` Ta có ` 2019 E = (2019^(2020) +2019)/(2019^(2020) +1) = 1 + 2018/(2019^(2020) +1)` ` 2019 F = ( 2019^(2021) + 2019)/(2019^(2021)+1) = 1 + 2018/(2019^(2021)+ 1)` Vì ` 2019^(2020) < 2019^(2021) => 2019^(2020) +1 < 2019^(2021) +1` ` => 2018/(2019^(2020) +1) > 2018/(2019^(2021)+ 1)` ` => 1 + 2018/(2019^(2020) +1) > 1 + 2018/(2019^(2021)+ 1)` ` => 2019 E > 2019 F` `=> E > F` Vậy ` E > F` Bình luận
Ta có: $E=\dfrac{2019^{2019}+1}{2019^{2020}+1}$ $\to 2019E=\dfrac{2019^{2020}+2019}{2019^{2020}+1}$ $=\dfrac{2019^{2020}+1+2018}{2019^{2020}+1}$ $=1+\dfrac{2018}{2019^{2020}+1}$ Lại có: $F=\dfrac{2019^{2020}+1}{2019^{2021}+1}$ $\to 2019F=\dfrac{2019^{2021}+2019}{2019^{2021}+1}$ $=\dfrac{2019^{2021}+1+2018}{2019^{2021}+1}$ $=1+\dfrac{2018}{2019^{2021}+1}$ Nhận thấy: $\dfrac{2018}{2019^{2020}+1}>\dfrac{2018}{2019^{2021}+1}$ $\to 1+\dfrac{2018}{2019^{2020}+1}>1+\dfrac{2018}{2019^{2021}+1}$ $↔2019E>2019F$ $\to E>F$ Bình luận
` E = (2019^(2019) +1)/(2019^(2020) +1`
` F = (2019^(2020) +1)/(2019^(2021) +1`
Ta có ` 2019 E = (2019^(2020) +2019)/(2019^(2020) +1) = 1 + 2018/(2019^(2020) +1)`
` 2019 F = ( 2019^(2021) + 2019)/(2019^(2021)+1) = 1 + 2018/(2019^(2021)+ 1)`
Vì ` 2019^(2020) < 2019^(2021) => 2019^(2020) +1 < 2019^(2021) +1`
` => 2018/(2019^(2020) +1) > 2018/(2019^(2021)+ 1)`
` => 1 + 2018/(2019^(2020) +1) > 1 + 2018/(2019^(2021)+ 1)`
` => 2019 E > 2019 F`
`=> E > F`
Vậy ` E > F`
Ta có:
$E=\dfrac{2019^{2019}+1}{2019^{2020}+1}$
$\to 2019E=\dfrac{2019^{2020}+2019}{2019^{2020}+1}$
$=\dfrac{2019^{2020}+1+2018}{2019^{2020}+1}$
$=1+\dfrac{2018}{2019^{2020}+1}$
Lại có:
$F=\dfrac{2019^{2020}+1}{2019^{2021}+1}$
$\to 2019F=\dfrac{2019^{2021}+2019}{2019^{2021}+1}$
$=\dfrac{2019^{2021}+1+2018}{2019^{2021}+1}$
$=1+\dfrac{2018}{2019^{2021}+1}$
Nhận thấy:
$\dfrac{2018}{2019^{2020}+1}>\dfrac{2018}{2019^{2021}+1}$
$\to 1+\dfrac{2018}{2019^{2020}+1}>1+\dfrac{2018}{2019^{2021}+1}$
$↔2019E>2019F$
$\to E>F$