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\(10A=\dfrac{10^{2023}+10}{10^{2023}+1}=1+\dfrac{9}{10^{2023}+1}\)
\(10B=\dfrac{10^{2022}+10}{10^{2022}+1}=1+\dfrac{9}{10^{2022}+1}\)
2023>2022
=>10^2023+1>10^2022+1
=>10A<10B
=>A<B
A = \(\dfrac{2022}{50^{10}}\) + \(\dfrac{2022}{50^8}\)
A = \(\dfrac{2022}{50^{10}}\) + \(\dfrac{2021}{50^8}\) + \(\dfrac{1}{50^8}\)
B = \(\dfrac{2023}{50^{10}}\) + \(\dfrac{2021}{5^8}\) = \(\dfrac{2022}{50^{10}}\) + \(\dfrac{1}{50^{10}}\) + \(\dfrac{2021}{50^8}\)
Vì: \(\dfrac{1}{50^{10}}\) < \(\dfrac{1}{50^8}\) nên \(\dfrac{2022}{50^{10}}\) + \(\dfrac{2021}{50^8}\) + \(\dfrac{1}{50^{10}}\) < \(\dfrac{2022}{50^{10}}\) + \(\dfrac{2021}{50^8}\) + \(\dfrac{1}{50^8}\)
Vậy A > B
a) 2021 + 2022 + 2023 + 2024 + 2025 + 2026 + 2027 + 2028 + 2029
= (2021 + 2029) + (2022 + 2028) + (2023 + 2027) + (2024 + 2026) + 2025
= 4050 + 4050 + 4050 + 4050 + 2025
= 4050.4 + 2025
= 16 200 + 2025
= 18 225
b)
30.40.50.60 = 3.10.4.10.5.10.6.10 = 3.4.5.6.10000 = 3.20.6.10000 = 3.2.6.10.10000 = 36.100000 = 3600000
\(\dfrac{1}{10}A=\dfrac{10^{2023}+5}{10^{2023}+50}=1-\dfrac{45}{10^{2023}+50}\)
\(\dfrac{1}{10}B=\dfrac{10^{2022}+5}{10^{2022}+50}=1-\dfrac{45}{10^{2022}+50}\)
10^2023+50>10^2022+50
=>-45/10^2023+50<-45/10^2020+50
=>1/10A<1/10B
=>A<B
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = (\(\dfrac{2021}{2}+1\))+(\(\dfrac{2020}{3}+1\))+....+(\(\dfrac{1}{2022}+1\))
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = \(\dfrac{2023}{2}\)+\(\dfrac{2023}{3}\)+....+ \(\dfrac{2023}{2022}\)
(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\)). x = 2023.( \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\))
vậy x= 2023
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
Đặt tổng trên là A
\(\dfrac{2A}{7}=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2021.2023}=\)
\(=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{2023-2021}{2021.2023}=\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2023}=\)
\(=\dfrac{2022}{2023}\Rightarrow A=\dfrac{7.2022}{2.2023}\)
\(A=\dfrac{7}{1.3}+\dfrac{7}{3.5}+\dfrac{7}{5.7}+...+\dfrac{7}{2021.2023}\\ \Rightarrow\dfrac{2}{7}A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2021.2023}\\ \Rightarrow\dfrac{2}{7}A=2-\dfrac{2}{3}+\dfrac{2}{3}-\dfrac{2}{5}+\dfrac{2}{5}-\dfrac{2}{7}+...+\dfrac{2}{2021}-\dfrac{2}{2023}\\ \Rightarrow\dfrac{2}{7}A=2-\dfrac{2}{2023}=\dfrac{4044}{2023}\Rightarrow A=\dfrac{2022}{289}\)
=-1/10(-21,22+11,22)+2021/2023(2025-2)
=2021-1/10*(-10)
=2021+1=2022