(7,5)^2022×1/72^2023×(-0,4)^2022×(-24)^2099
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So sánh
A = \(\dfrac{2022^{2023}+1}{2022^{2024}+1}\) và B = \(\dfrac{2022^{2022}+1}{2022^{2023}+1}\)
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
\(\dfrac{2022\times2023-1}{2023\times2021+2022}\)
= \(\dfrac{\left(2021+1\right)\times2023-1}{2023\times2021+2022}\)
= \(\dfrac{2023\times2021+2023-1}{2023\times2021+2022}\)
= \(\dfrac{2023\times2021+2022}{2023\times2021+2022}\)
= 1
Lời giải:
\(A=2.2022^{2023}+2(1^{2023}+2^{2023}+3^{2023}+...+1010^{2023}+1011^{2023}+1012^{2023}+...+2021^{2023})\)
\(=2.2022^{2023}+2[(1^{2023}+2021^{2023})+(2^{2023}+2019^{2023})+...+(1010^{2023}+1012^{2023})+1011^{2023}]\)
\(=2.2022^{2023}+2.1011^{2023}+2[(1^{2023}+2021^{2023})+(2^{2023}+2019^{2023})+...+(1010^{2023}+1012^{2023})]\)
Dễ thấy: $2.2022^{2023}\vdots 2022; 2.1011^{2023}=2022.1011^{2023}\vdots 2022$
Đối với biểu thức trong ngoặc vuông thì: Nhớ rằng với mọi $n$ lẻ thì $a^n+b^n\vdots a+b$ nên $1^{2023}+2021^{2023}\vdots 2022; 2^{2023}+2019^{2023}\vdots 2022;...; 1010^{2023}+1012^{2023}\vdots 2022$
$\Rightarrow 2[(1^{2023}+2021^{2023})+(2^{2023}+2019^{2023})+....+(1010^{2023}+1012^{2023})]\vdots 2022$
Do đó $A\vdots 2022$
\(\left(7,5\right)^{2022}\cdot\left(-0,4\right)^{2022}\cdot\dfrac{1}{72^{2023}}\cdot\left(-24\right)^{2099}\)
\(=\left(-3\right)^{2022}\cdot\dfrac{1}{24^{2023}\cdot3^{2023}}\cdot\left(-24\right)^{2099}\)
\(=\dfrac{3^{2022}}{3^{2023}}\cdot24^{76}\cdot\left(-1\right)=\dfrac{-24^{76}}{3}=\dfrac{-3^{76}\cdot8^{76}}{3}\)
\(=-3^{75}\cdot8^{76}\)
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