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

 2018^2019+1/2018^2020+1 bé hơn 2018^2020+1/2018^2021+1 

x=2020 nên x+1=2021

\(P\left(x\right)=x^{2021}-x^{2020}\left(x+1\right)+x^{2019}\left(x+1\right)-....+x\left(x+1\right)-2020\)

\(=x^{2021}-x^{2021}-x^{2020}+x^{2020}-...+x^2+x-2020\)

=x-2020=0

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

\(A=\dfrac{2020^{2018}-1}{2020^{2019}+2019}\)

\(B=\dfrac{2020^{2019}+1}{2020^{2020}+2019}\)

Ta có :

\(A-B=\dfrac{2020^{2018}-1}{2020^{2019}+2019}-\dfrac{2020^{2019}+1}{2020^{2020}+2019}\)

\(\Rightarrow A-B=\dfrac{\left(2020^{2018}-1\right)\left(2020^{2020}+2019\right)-\left(2020^{2019}+2019\right)\left(2020^{2019}+1\right)}{\left(2020^{2019}+2019\right)\left(2020^{2020}+2019\right)}\)

\(\Rightarrow A-B=\dfrac{2020^{4038}+2019.2020^{2018}-2020^{2020}-2019-2020^{4038}-2020^{2019}-2019.2020^{2018}-2029}{\left(2020^{2019}+2019\right)\left(2020^{2020}+2019\right)}\)

\(\Rightarrow A-B=\dfrac{-\left(2020^{2020}+2020^{2019}+2.2019\right)}{\left(2020^{2019}+2019\right)\left(2020^{2020}+2019\right)}\)

mà \(\left\{{}\begin{matrix}-\left(2020^{2020}+2020^{2019}+2.2019\right)< 0\\\left(2020^{2019}+2019\right)\left(2020^{2020}+2019\right)>0\end{matrix}\right.\)

\(\Rightarrow A-B< 0\)

\(\Rightarrow A< B\)

Vậy ta được \(A< B\)

Ta có :\(\frac{x+4}{2018}+\frac{x+3}{2019}=\frac{x+2}{2020}+\frac{x+1}{2021}\)

=> \(\left(\frac{x+4}{2018}+1\right)+\left(\frac{x+3}{2019}+1\right)=\left(\frac{x+2}{2020}+1\right)+\left(\frac{x+1}{2021}+1\right)\)

=> \(\frac{x+2022}{2018}+\frac{x+2022}{2019}=\frac{x+2022}{2020}+\frac{x+2022}{2021}\)

=> \(\frac{x+2022}{2018}+\frac{x+2022}{2019}-\frac{x+2022}{2020}-\frac{x+2022}{2021}=0\)

=> \(\left(x+2022\right)\left(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\right)=0\)

Vì \(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\ne0\)

=> x + 2022 = 0

=> x = -2022

Vậy x = -2022

\(\frac{x+4}{2018}+\frac{x+3}{2019}=\frac{x+2}{2020}+\frac{x+1}{2021}\)  

\(\frac{x+4}{2018}+1+\frac{x+3}{2019}+1=\frac{x+2}{2020}+1+\frac{x+1}{2021}+1\) 

\(\frac{x+4}{2018}+\frac{2018}{2018}+\frac{x+3}{2019}+\frac{2019}{2019}=\frac{x+2}{2020}+\frac{2020}{2020}+\frac{x+1}{2021}+\frac{2021}{2021}\)   

\(\frac{x+2022}{2018}+\frac{x+2022}{2019}=\frac{x+2022}{2020}+\frac{x+2022}{2021}\)   

\(\frac{x+2022}{2018}+\frac{x+2022}{2019}-\frac{x+2022}{2020}-\frac{x+2022}{2021}=0\)   

\(\left(x+2022\right)\left(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\right)=0\)   

\(x+2022=0\left(\frac{1}{2018}+\frac{1}{2019}-\frac{1}{2020}-\frac{1}{2021}\ne0\right)\)   

\(x=0-2022\) 

\(x=-2022\)

Đề bài yêu cầu gì?

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