M=0 vì trong bt có \(1-\frac{2018}{1018}=1-1=0\)

2019 + 2017 + 2015 + .... + 1003 + 1001 - 2018 - 2016 - 2014 -.... - 1002 - 1000

=(2019-2018)+(2017-2016)+....+(1001-1000)

=1+1+...+1

= 510

\(A=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2018\times2019}\)

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

\(A=\left(\dfrac{2020}{2019}-\dfrac{1}{2019}\right)-\left(\dfrac{2019}{2018}-\dfrac{1}{2018}\right)\)

\(A=\left(\dfrac{2020-1}{2019}\right)-\left(\dfrac{2019-1}{2018}\right)\)

\(A=1-1\)

\(A=0.\)

\(A=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2018\times2019}\)

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

\(A=\left(\dfrac{2020}{2019}-\dfrac{1}{2019}\right)-\left(\dfrac{2019}{2018}-\dfrac{1}{2018}\right)\)

\(A=\dfrac{2019}{2019}-\dfrac{2018}{2018}\)

\(A=1-1\)

\(A=0\)

Ta có: B = (2018 + 2019)/(2019 + 2020) = (2018 + 2019)/4039 = 2018/4039 + 2019/4039
Ta thấy : 2018/2019 > 2018/4039
            2019/2020 > 2019/4039
=> 2018/2019 + 2019/2020 > 2018/4039 > 2019/4039
=> 2018/2019 + 2019/2020 > (2018 + 2019)/(2019 + 2020)
=> A  > B

\(A=\frac{2018^{2019}-1}{2018^{2019}+1}=\frac{2018^{2019}+1-2}{2018^{2019}+1}=\frac{2018^{2019}+1}{2018^{2019}+1}-\frac{2}{2018^{2019}+1}=1-\frac{2}{2018^{2019}+1}\)

\(B=\frac{2018^{2019}}{2018^{2019}+2}=\frac{2018^{2019}+2-2}{2018^{2019}+2}=\frac{2018^{2019}+2}{2018^{2019}+2}-\frac{2}{2018^{2019}+2}=1-\frac{2}{2018^{2019}+2}\)

Ta có: \(\frac{2}{2018^{2019}+1}>\frac{2}{2018^{2019}+2}\)

\(\Rightarrow1-\frac{2}{2018^{2019}+1}< 1-\frac{2}{2018^{2019}+2}\)

\(\Rightarrow A< B\)

Vậy .....

\(M=\frac{2018^{2018}+1}{2019^{2019}+1}\)

\(\Leftrightarrow2M=1+\frac{2017}{2018^{2019}+1}\)

\(N=\frac{2018^{2019}-2}{2018^{2020}-2}\)

\(\Leftrightarrow2N=1-\frac{4034}{2018^{2020}-2}\)

Nhận thấy :  \(1+\frac{2017}{2018^{2019}+1}>1-\frac{4034}{2018^{2020}-2}\Leftrightarrow2M>2N\Leftrightarrow M>N\)

Từ đề bài, ta suy ra:

So sánh hai biểu thức

\(M=\left(2018^{2018}+1\right)\cdot\left(2018^{2020}-2\right)\)(1)

\(N=\left(2018^{2019}-2\right)\cdot\left(2018^{2019}+1\right)\)(2)

Xét biểu thức M và N, ta suy ra:

\(M=\left(2018^{2019}-2017\right)\cdot\left(2019^{2019}+2016\right)\)

\(N=\left(2018^{2019}-2017\right)\cdot\left(2018^{2018}-2016\right)\)

Nhận thấy (2019²⁰¹⁹+2016)>(2018²⁰¹⁸-2016) nên M>N

Vậy M>N.

P/s:Mình đây không phải top 10 tuần nên bài có thể sai sót, mong bạn tham khảo:)))

2018 A = 2018 - 2018^2 + 2018^3 +...- 2018^2018 + 2018^2019

=> A + 2018 A = 1 +2018^2019

=> 2019 A = 1 + 2018^2019 

=> 2019 A - 1 = 2018^2019 

=> 2019 A -1 là 1 lũy thừa của 2018

\(\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{1}{2019}x2018\)
\(=\dfrac{2020}{2019}-\dfrac{2019}{2018}+\dfrac{2018}{2019}=2-\dfrac{2019}{2018}=\dfrac{2017}{2018}\)