So sánh: \(\dfrac{2022}{\sqrt{2023}}\) + \(\dfrac{2023}{\sqrt{2022}}\) và \(\sqrt{2022}\) + \(\sqrt{2023}\)
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 \(A_2>A_1\).
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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
A=\(\dfrac{2022^{2022}+1}{2022^{2023}+1}\) ; B= \(\dfrac{2022^{2023}+1}{2022^{2024}+1}\) So sánh A và B
SOS! Mình đang cần giúp nhanh ạ vì mai thi rồi cảm ơn nhiều
So sánh
a)17/20 và 18/19 b)19/18 và 2023/2022
c)13/17 và 135/175 d)53/63 và 535/636
e)13/15 và 22/25 \(\dfrac{2023}{2023^2+1}và\dfrac{2022}{2022^2+1}\)
a) \(\dfrac{17}{20}< \dfrac{18}{20}< \dfrac{18}{19}\Rightarrow\dfrac{17}{20}< \dfrac{18}{19}\)
b) \(\dfrac{19}{18}>\dfrac{19+2024}{18+2024}=\dfrac{2023}{2022}\Rightarrow\dfrac{19}{18}>\dfrac{2023}{2022}\)
c) \(\dfrac{135}{175}=\dfrac{27}{35}\)
\(\dfrac{13}{17}=\dfrac{26}{34}< \dfrac{26+1}{34+1}=\dfrac{27}{35}\)
\(\Rightarrow\dfrac{13}{17}< \dfrac{135}{175}\)
So sánh 2 phân số
A = \(\dfrac{2022^{2022}+1}{2022^{2021}+1}\) ; B = \(\dfrac{2022^{2023}+1}{2021^{2022}+1}\)
So sánh A= \(\dfrac{10^{2023}+5}{10^{2022}+5}\) và B= \(\dfrac{10^{2022}+5}{10^{2022}+5}\)
A và B có phần mẫu số bằng nhau mà tử A có 10^2023 lớn hơn B có 10^2022 => A > B
10^2023>10^2022
=>10^2023+5>10^2022+5
=>A>B
So sánh:\(\dfrac{2023}{2022}\) và \(\dfrac{2022}{2021}\)
Bạn nào nhanh mình tick cho!
2023/2022=1+1/2022
2022/2021=1+1/2021
mà 2022>2021
nên 2023/2022<2022/2021
so sánh
\(\dfrac{10^{2023}-3}{10^{2024}-3}\)
và
\(\dfrac{10^{2022}+1}{10^{2023}+1}\)
Ta có :
\(\dfrac{10^{2023}}{10^{2024}}=\dfrac{10^{2022}}{10^{2023}}\)
mà \(\dfrac{10^{2023}}{10^{2024}}>\dfrac{10^{2023}-3}{10^{2024}-3}\)
\(\dfrac{10^{2022}}{10^{2023}}< \dfrac{10^{2022}+1}{10^{2023}+1}\)
\(\Rightarrow\dfrac{10^{2023}-3}{10^{2024}-3}< \dfrac{10^{2022}+1}{10^{2023}+1}\)
So sánh A= \(\dfrac{10^{2023}+5}{10^{2022}+5}\) và B=\(\dfrac{10^{2022}+5}{10^{2021}+5}\)
\(\dfrac{1}{10}A=\dfrac{10^{2023}+5}{10^{2023}+50}=1-\dfrac{45}{10^{2023}+50}\)
\(\dfrac{1}{10}B=\dfrac{10^{2022}+5}{10^{2022}+50}=1-\dfrac{45}{10^{2022}+50}\)
10^2023+50>10^2022+50
=>-45/10^2023+50<-45/10^2020+50
=>1/10A<1/10B
=>A<B