1

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

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

2

\(M=\frac{100^{101}+1}{100^{100}+1}< \frac{100^{101}+1+99}{100^{100}+1+99}=\frac{100^{101}+100}{100^{100}+100}=\frac{100\left(100^{100}+1\right)}{100\left(100^{99}+1\right)}\)

\(=\frac{100^{100}+1}{100^{99}+1}=N\)

\(A=\frac{100^{2017}+1}{100^{2018}+1}\)

\(\Rightarrow100A=\frac{100\cdot\left[100^{2017}+1\right]}{100^{2018}+1}\)

\(\Rightarrow100A=\frac{100^{2018}+100}{100^{2018}+1}\)

\(\Rightarrow100A=\frac{100^{2018}+1+99}{100^{2018}+1}\)

\(\Rightarrow100A=1+\frac{99}{100^{2018}+1}\)

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

\(\Rightarrow100B=\frac{100\cdot\left[100^{2018}+1\right]}{100^{2019}+1}\)

\(\Rightarrow100B=\frac{100^{2019}+100}{100^{2019}+1}\)

\(\Rightarrow100B=\frac{100^{2019}+1+99}{100^{2019}+1}\)

\(\Rightarrow100B=1+\frac{99}{100^{2019}+1}\)

Tự so sánh

\(A=\frac{100^{2017}+1}{100^{2018}+1}\)

\(\Rightarrow100A=\frac{100^{2018}+100}{100^{2018}+1}\)

\(\Rightarrow100A=\frac{100^{2018}+1+99}{100^{2018}+1}\)

\(\Rightarrow100A=\frac{100^{2018}+1}{100^{2018}+1}+\frac{99}{100^{2018}+1}\)

\(\Rightarrow100A=1+\frac{99}{100^{2018}+1}\)(1)

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

\(\Rightarrow100B=\frac{100^{2019}+100}{100^{2019}+1}\)

\(\Rightarrow100B=\frac{100^{2019}+1+99}{100^{2019}+1}\)

\(\Rightarrow100B=\frac{100^{2019}+1}{100^{2019}+1}+\frac{99}{100^{2019}+1}\)

\(\Rightarrow100B=1+\frac{99}{100^{2019}+1}\)(2)

Từ (1) và (2) suy ra 100A > 100B hay A > B

\(B=\frac{100^{2018}+1}{100^{2019}+1}< \frac{100^{2018}+1+99}{100^{2019}+1+99}=\frac{100^{2018}+100}{100^{2019}+100}=\frac{100\left(100^{2017}+1\right)}{100\left(100^{2018}+1\right)}=\frac{100^{2017}+1}{100^{2018}+1}=A\)

\(\Rightarrow B< A\)

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 .....

B= 1/1.2+1/2.3+...+1/2019.2020

B=1/1-1/2+1/2-1/3+...+1/2019-1/2020

B=1-1/2020=2020/2020-1/2020=2019/2020

        \(2018^{100}+2018^{99}\)

\(=2018^{99}.\left(2018+1\right)\)

\(=2018^{99}.2019\)\(< 2019^{99}.2019=2019^{100}\)

\(\Rightarrow2018^{100}+2018^{99}< 2019^{100}\)

                          Vậy \(2018^{100}+2018^{99}< 2019^{100}\)

                                                      ~~Hok tốt~~

\(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1.\) 

Với  :   \(a=2^{2018};.b=3^{2019};,c=5^{2020}.\) 

Và   :   \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2019.2020}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\Leftrightarrow\) 

             \(B=1-\frac{1}{2020}< 1< A\)