so sánh 2022^45 -2022^44 và 2022^44-2022^43
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(2022A=2022+2022^2+2022^3+2022^4+...+2022^{2018}\)
\(2021A=2022A-A=2022^{2018}-1\Rightarrow A=\dfrac{2022^{2018}-1}{2021}\)
\(\Rightarrow A< B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(2023^{2022}=2023\cdot2023^{2021}\)
\(2022^{2022}+2022^{2021}=2022^{2021}\cdot\left(2022+1\right)=2023\cdot2022^{2021}\)
Mà: \(2023>2022\)
\(\Rightarrow2023^{2021}>2022^{2021}\)
\(\Rightarrow2023^{2021}\cdot2023>2022^{2021}\cdot2023\)
\(\Rightarrow2023^{2022}>2022^{2022}+2022^{2021}\)
Vậy: ...
![](https://rs.olm.vn/images/avt/0.png?1311)
So sánh
A = \(\dfrac{2022^{2023}+1}{2022^{2024}+1}\) và B = \(\dfrac{2022^{2022}+1}{2022^{2023}+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Trước hết ta phải chứng minh \(\dfrac{a}{b}< \dfrac{a+1}{b+1}\) (a, b ϵ N; a < b).
Thật vậy, \(\dfrac{a}{b}=\dfrac{a\left(b+1\right)}{b\left(b+1\right)}=\dfrac{a+ab}{b^2+b}\) và \(\dfrac{a+1}{b+1}=\dfrac{\left(a+1\right)b}{\left(b+1\right)b}=\dfrac{ab+b}{b^2+b}\).
Mà theo giả thuyết là a < b nên \(\dfrac{a+ab}{b^2+b}< \dfrac{ab+b}{b^2+b}\), suy ra \(\dfrac{a}{b}< \dfrac{a+1}{b+1}\) (a, b ϵ N; a < b).
Từ đây ta có:
\(B=\dfrac{2022^{2022}+1}{2022^{2023}+1}=\dfrac{2022^{2023}+2022}{2022^{2024}+2022}=\dfrac{2022^{2023}+2021+1}{2022^{2024}+2021+1}\)
Đặt \(A_1=\dfrac{2022^{2023}+2}{2022^{2024}+2}=\dfrac{2022^{2023}+1+1}{2022^{2024}+1+1}\), rõ ràng \(A_1>A\).
Đặt \(A_2=\dfrac{2022^{2023}+3}{2022^{2024}+3}=\dfrac{2022^{2023}+2+1}{2022^{2024}+2+1}\), rõ ràng \(A_2>A_1\).
...
Đặt \(A_{2020}=\dfrac{2022^{2023}+2021}{2022^{2024}+2021}=\dfrac{2022^{2023}+2020+1}{2022^{2024}+2020+1}\), rõ ràng \(A_{2020}>A_{2019}\) và \(B>A_{2020}\).
Suy ra \(B>A_{2020}>A_{2019}>...>A_2>A_1>A\). Vậy A < B.
Ta có A = \(\dfrac{2022^{2023}}{2022^{2024}}=\dfrac{1}{2022}\) ; B = \(\dfrac{2022^{2022}}{2022^{2023}}=\dfrac{1}{2022}\)
Mà \(\dfrac{1}{2022}=\dfrac{1}{2022}\)
Vậy A = B
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{2022}\cdot A=\dfrac{2022^{100}+1}{2022^{100}+100}=1-\dfrac{99}{2022^{100}+100}\)
\(\dfrac{1}{2022}B=\dfrac{2022^{101}+1}{2022^{101}+100}=1-\dfrac{9}{2022^{101}+100}\)
2022^100+100<2022^101+100
=>-99/2022^100+100<-99/2022^101+100
=>A<B
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: 202245-202244=202244.(202245-44-1)=202244.(2022-1)=202244.2021
202244-202243=202243.(202244-43-1)=202243.(2022-1)=202243.2021
Vì 202244.2021>202243.2021 nên 202245-202244>202244-202243
\(2022^{45}-2022^{44}=2022^{44}\cdot2021\)
\(2022^{44}-2022^{43}=2022^{43}\cdot2021\)
mà 44>43
nên \(2022^{45}-2022^{44}>2022^{44}-2022^{43}\)