Xét 2017 /2018 và 2018/2019

1-2017/2018=1/2018

1-2018/2019=1/2019

mà 1/2018>1/2019=>2017/2018<2018/2019

Tương tự có:2020/2019>2021/2020

=>2017/2018+2010/2019<2018/2019+2021/2020

\(x=\frac{2019^{2020}+1}{2019^{2019}+1}>\frac{2019^{2020}+1+2018}{2019^{2019}+1+2018}=\frac{2019^{2020}+2019}{2019^{2019}+2019}=\frac{2019\left(2019^{2019}+1\right)}{2019\left(2019^{2018}+1\right)}=\frac{2019^{2019}+1}{2019^{2018}+1}\)(1)

\(y=\frac{2019^{2019}+2020}{2019^{2018}+2020}< \frac{2019^{2019}+2020-2019}{2019^{2018}+2020-2019}=\frac{2019^{2019}+1}{2019^{2018}+1}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow x>y\)

khó quá hè

a)2017²⁰¹⁸=...7⁸=...4

2018²⁰¹⁹=...8⁹=...8

2019²⁰²⁰=...9⁰=...0

2020²⁰²¹=...0

Vì 4+8+0+8=...0

Vậy A chia hết cho 10

a) Ta có : \(\frac{-60}{12}=-5=-\frac{25}{5}\)

\(-0,8=-\frac{8}{10}=-\frac{4}{5}\)

Mà -25 < -4 nên \(\frac{-25}{5}< \frac{-4}{5}\)=> \(\frac{-60}{12}< -0,8\)

b) Ta có : \(\frac{2020}{2019}=1+\frac{1}{2019}\)

\(\frac{2021}{2020}=1+\frac{1}{2020}\)

Vì \(\frac{1}{2019}>\frac{1}{2020}\)nên \(\frac{2020}{2019}>\frac{2021}{2020}\)

c) \(\frac{10^{2018}+1}{10^{2019}+1}=\frac{10\left(10^{2018}+1\right)}{10^{2019}+1}=\frac{10^{2019}+10}{10^{2019}+1}=\frac{10^{2019}+1+9}{10^{2019}+1}=1+\frac{9}{10^{2019}+1}\)(1)

\(\frac{10^{2019}+1}{10^{2020}+1}=\frac{10\left(10^{2019}+1\right)}{10^{2020}+1}=\frac{10^{2020}+10}{10^{2020}+1}=\frac{10^{2020}+1+9}{10^{2020}+1}=1+\frac{9}{10^{2020}+1}\)(2)

Đến đây tự so sánh rồi nhé

Ta có : \(\hept{\begin{cases}\frac{2019}{2020}< 1\\\frac{2018}{2018}=1\end{cases}\Rightarrow\frac{2019}{2020}< \frac{2018}{2018}}\)

Ta có : 

\(\frac{2019}{2020}< 1\)

\(\frac{2018}{2018}=1\)

\(\Rightarrow\frac{2019}{2020}< \frac{2018}{2018}\)

#Riin