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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
2020/2021<1
2021/2022<1
2022/2023<1
2023/2020=1+1/2020+1/2020+1/2020>1+1/2021+1/2022+1/2023
=>B>2020/2021+2021/2022+2022/2023+1/2021+1/2022+1/2023+1=4
Ta có:
\(10A=\dfrac{10\left(10^{2020}+1\right)}{10^{2021}+1}=\dfrac{10^{2021}+10}{10^{2021}+1}=1+\dfrac{9}{10^{2021}+1}\)
\(10B=\dfrac{10\left(10^{2021}+1\right)}{10^{2022}+1}=\dfrac{10^{2022}+10}{10^{2022}+1}=1+\dfrac{9}{10^{2022}+1}\)
⇒ \(10A>10B\) ( vì \(\dfrac{9}{10^{2021}+1}>\dfrac{9}{10^{2022}+1}\) )
Suy ra: \(A>B\)
Ta có: \(B=2020.2021.2022=\left(2021-1\right).\left(2021+1\right).2021=\left(2021-1\right)^2.2021< 2021^2.2021=A\)
Ta có: \(A=\frac{2020}{2021}+\frac{2021}{2022}\)
\(\Rightarrow A=\frac{2021}{2021}-\frac{1}{2021}+\frac{2022}{2022}-\frac{1}{2022}\)
\(\Rightarrow A=1-\frac{1}{2021}+1-\frac{1}{2022}\)
\(\Rightarrow A=1+1-\frac{1}{2021}-\frac{1}{2022}\)
\(\Rightarrow A=2-\frac{1}{2021}-\frac{1}{2022}\)
\(\Rightarrow A=2-\frac{1}{2021\cdot2022}\)
\(B=\frac{2020+2021}{2021+2022}\)
\(\Rightarrow B=\frac{2021+2022}{2021+2022}-\frac{2}{2021+2022}\)
\(\Rightarrow B=1-\frac{2}{2021+2022}\)
\(\Rightarrow B=1-\frac{2}{4043}\)
Vậy ta sẽ so sánh:
\(1-\frac{1}{2021\cdot2022};\frac{2}{4043}\)
Vì \(2021\cdot2022>4043\)nên \(\frac{1}{2021\cdot2022}< \frac{2}{4043}\)vậy \(1-\frac{1}{2021\cdot2022}>\frac{2}{4043}\)
\(\Rightarrow\frac{2020}{2021}+\frac{2021}{2022}>\frac{2020+2021}{2021+2022}\)
\(\Rightarrow A>B\)