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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
Ta có : A = \(\frac{10^{2020}+1}{10^{2021}+1}\)
=> 10A = \(\frac{10^{2021}+10}{10^{2021}+1}=1+\frac{9}{10^{2021}+1}\)
Lại có : \(B=\frac{10^{2021}+1}{10^{2022}+1}\)
=> \(10B=\frac{10^{2022}+10}{10^{2022}+1}=1+\frac{9}{10^{2022}+1}\)
Vì \(\frac{9}{10^{2022}+1}< \frac{9}{10^{2021}+1}\)
=> \(1+\frac{9}{10^{2022}+1}< 1+\frac{9}{10^{2022}+1}\)
=> 10B < 10A
=> B < A
b) Ta có : \(\frac{2019}{2020+2021}< \frac{2019}{2020}\)
Lại có : \(\frac{2020}{2020+2021}< \frac{2020}{2021}\)
=> \(\frac{2019}{2020+2021}+\frac{2020}{2020+2021}< \frac{2019}{2020}+\frac{2020}{2021}\)
=> \(\frac{2019+2020}{2020+2021}< \frac{2019}{2020}+\frac{2020}{2021}\)
=> B < A
\(10A=\dfrac{10^{2023}+10}{10^{2023}+1}=1+\dfrac{9}{10^{2023}+1}\)
\(10B=\dfrac{10^{2022}+10}{10^{2022}+1}=1+\dfrac{9}{10^{2022}+1}\)
2023>2022
=>10^2023+1>10^2022+1
=>10A<10B
=>A<B
\(10A=\dfrac{10^{2023}+10}{10^{2023}+1}=1+\dfrac{9}{10^{2023}+1}\)
\(10B=\dfrac{10^{2022}+10}{10^{2022}+1}=1+\dfrac{9}{10^{2022}+1}\)
mà 10^2023+1>10^2022+1
nên A<B
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
\(\dfrac{1}{10}A=\dfrac{10^{2023}+5}{10^{2023}+50}=1-\dfrac{45}{10^{2023}+50}\)
\(\dfrac{1}{10}B=\dfrac{10^{2022}+5}{10^{2022}+50}=1-\dfrac{45}{10^{2022}+50}\)
10^2023+50>10^2022+50
=>-45/10^2023+50<-45/10^2020+50
=>1/10A<1/10B
=>A<B
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