a: \(98^{10}\cdot A=\dfrac{98^{98}+98^{10}}{98^{98}+1}=1+\dfrac{98^{10}-1}{98^{98}+1}\)

\(98^{10}\cdot B=\dfrac{98^{99}+98^{10}}{98^{99}+1}=1+\dfrac{98^{10}-1}{98^{99}+1}\)

98^88+1>98^99+1

=>A<B

b: \(\dfrac{1}{2022^2}\cdot C=\dfrac{2022^{2023}+1}{2022^{2023}+2022^2}=1+\dfrac{1-2022^2}{2022^{2023}+2022^2}\)

\(\dfrac{1}{2022^2}\cdot D=\dfrac{2022^{2021}+1}{2022^{2021}+2022^2}=1+\dfrac{1-2022^2}{2022^{2021}+2022^2}\)

2022^2023>2022^2021

=>2022^2023+2022^2>2022^2021+2022^2

=>\(\dfrac{2022^2-1}{2022^{2023}+2022^2}< \dfrac{2022^2-1}{2022^{2021}+2022^2}\)

=>\(\dfrac{1-2022^2}{2022^{2023}+2022^2}>\dfrac{1-2022^2}{2022^{2021}+2022^2}\)

=>C>D

\(10A=\dfrac{10^{2023}+10}{10^{2023}+1}=1+\dfrac{9}{10^{2023}+1}\)

\(10B=\dfrac{10^{2022}+10}{10^{2022}+1}=1+\dfrac{9}{10^{2022}+1}\)

2023>2022

=>10^2023+1>10^2022+1

=>10A<10B

=>A<B

giúp vơi

khó difficult

B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + \(\dfrac{2022}{1}\)

B = \(\dfrac{1}{2002}\) + \(\dfrac{2}{2021}\) + \(\dfrac{3}{2020}\)+...+ \(\dfrac{2021}{2}\) + 2022

B = 1 + ( 1 + \(\dfrac{1}{2022}\)) + ( 1 + \(\dfrac{2}{2021}\)) + \(\left(1+\dfrac{3}{2020}\right)\)+ ... + \(\left(1+\dfrac{2021}{2}\right)\) 

B = \(\dfrac{2023}{2023}\) + \(\dfrac{2023}{2022}\) + \(\dfrac{2023}{2021}\) + \(\dfrac{2023}{2020}\) + ...+ \(\dfrac{2023}{2}\) 

B = 2023 \(\times\) ( \(\dfrac{1}{2023}\) + \(\dfrac{1}{2022}\) + \(\dfrac{1}{2021}\) + \(\dfrac{1}{2020}\)+ ... + \(\dfrac{1}{2}\))

Vậy B > C 

Ta có: \(\frac{2022}{2021^2+k}\le\frac{2022}{2021^2}\) (với \(k\)là số tự nhiên bất kì) 

Ta có: 

\(A=\frac{2022}{2021^2+1}+\frac{2022}{2021^2+2}+...+\frac{2022}{2021^2+2021}\)

\(\le\frac{2022}{2021^2}+\frac{2022}{2021^2}+...+\frac{2022}{2021^2}=\frac{2022}{2021^2}.2021=\frac{2022}{2021}\)

Ta có: \(\frac{2022}{2021^2+k}>\frac{2022}{2021^2+2021}=\frac{2022}{2021.2022}=\frac{1}{2021}\)với \(k\)tự nhiên, \(k< 2021\)) 

Suy ra \(A=\frac{2022}{2021^2+1}+\frac{2022}{2021^2+2}+...+\frac{2022}{2021^2+2021}\)

\(>\frac{1}{2021}+\frac{1}{2021}+...+\frac{1}{2021}=\frac{2021}{2021}=1\)

Suy ra \(1< A\le\frac{2022}{2021}\)do đó \(A\)không phải là số tự nhiên. 

Nhỏ hơn

Ta có 2020/2021 <1

         2021/2022 <1

         2022/2023 <1

         2023/2024 <1

Suy ra A=(2021/2021+2021/2022 +2022/2023 +2023/2024) < (1+1+1+1)= 4

      Vậy A <4

Chúc bạn học tốt

\(\dfrac{2020}{2021}< 1\)

\(\dfrac{2021}{2022}< 1\)

\(\dfrac{2021}{2022}< 1\)

\(\dfrac{2023}{2024}< 1\)

Do đó: A<4

a: \(B=\dfrac{154}{155+156}+\dfrac{155}{155+156}\)

\(\dfrac{154}{155}>\dfrac{154}{155+156}\)

\(\dfrac{155}{156}>\dfrac{155}{155+156}\)

=>154/155+155/156>(154+155)/(155+156)

=>A>B

b: \(C=\dfrac{2021+2022+2023}{2022+2023+2024}=\dfrac{2021}{6069}+\dfrac{2022}{6069}+\dfrac{2023}{6069}\)

2021/2022>2021/6069

2022/2023>2022/2069

2023/2024>2023/6069

=>D>C