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

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

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 .....

Ta tính hiệu của M và T

ta có 

Hiệu của Mẫu và Tử của A là   2019^2019-1 - (2019^2018-1) = 2019^2019 - 2019^2018 = 2019^2019.2018

Hiệu của Mẫu và Tử của B là   2019^2019+1 - (2019^2018+1) = 2019^2019 - 2019^2018 = 2019^2019.2018

2 Hiệu trên bằng nhau nên A < B  

Ta có: \(C=\dfrac{2019-2018}{2019+2018}\)

\(\Leftrightarrow C=\dfrac{\left(2019-2018\right)\left(2019+2018\right)}{\left(2019+2018\right)^2}\)

\(\Leftrightarrow C=\dfrac{2019^2-2018^2}{\left(2019+2018\right)^2}\)

Ta có: \(\left(2019+2018\right)^2=2019^2+2018^2+2\cdot2019\cdot2018\)

\(2019^2+2018^2=2019^2+2018^2+0\)

Do đó: \(\left(2019+2018\right)^2>2019^2+2018^2\)

\(\Leftrightarrow\dfrac{2019^2-2018^2}{\left(2019+2018\right)^2}< \dfrac{2019^2-2018^2}{2019^2+2018^2}\)

\(\Leftrightarrow C< D\)

Bài toán : So sánh A và B

\(A=\frac{2018^{100}}{1+2018+2018^2+...+2018^{100}}\)

+) Ta có \(\frac{1}{A}=\frac{1+2018+2018^2+...+2018^{100}}{2018^{100}}\)

                     \(=\frac{1}{2018^{100}}+\frac{2018}{2018^{100}}+\frac{2018^2}{2018^{100}}+...+\frac{2018^{100}}{2018^{100}}\)

                      \(=\frac{1}{2018^{100}}+\frac{1}{2018^{99}}+\frac{1}{2018^{98}}+...+1\)

\(B=\frac{2019^{100}}{1+2019+2019^2+...+2019^{100}}\)

+) Ta có \(\frac{1}{B}=\frac{1+2019+2019^2+...+2019^{100}}{2019^{100}}\)

                     \(=\frac{1}{2019^{100}}+\frac{2019}{2019^{100}}+\frac{2019^2}{2019^{100}}+...+\frac{2019^{100}}{2019^{100}}\)

                      \(=\frac{1}{2019^{100}}+\frac{1}{2019^{99}}+\frac{1}{2019^{98}}+...+1\)

+) \(\frac{1}{2018^{100}}>\frac{1}{2019^{100}}\)

     \(\frac{1}{2018^{99}}>\frac{1}{2019^{99}}\)

     .....................................

     \(1=1\)

\(\Rightarrow\frac{1}{2018^{100}}+\frac{1}{2018^{99}}+\frac{1}{2018^{98}}+...+1>\frac{1}{2019^{100}}+\frac{1}{2019^{99}}+\frac{1}{2019^{98}}+...+1\)

\(\Rightarrow\frac{1}{A}>\frac{1}{B}\)

\(\Rightarrow A< B\)

Vậy \(A< B\)

Ta có : \(0< \frac{2017}{2018}< 1\) nên   \(\frac{2017}{2018}>\frac{2017+2019}{2018+2019}\)(1)

\(0< \frac{2018}{2019}< 1\) nên \(\frac{2018}{2019}>\frac{2018+2018}{2018+2019}\) (2)

Cộng vế theo vế 1 và 2 ta được : \(B=\frac{2017}{2018}+\frac{2018}{2019}>\frac{2017+2018+2018+2019}{2018+2019}=\frac{2017+2018}{2018 +2019}+1=A+1>A\)

Vậy B>A