so sánh A =2022/2023+2023/2024+2024/2025 với 3
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ND
1
NN
13 tháng 2 2023
\(A=\dfrac{2024^{2023}+1}{2024^{2024}+1}\)
\(2024A=\dfrac{2024^{2024}+2024}{2024^{2024}+1}=\dfrac{\left(2024^{2024}+1\right)+2023}{2024^{2024}+1}=\dfrac{2024^{2024}+1}{2024^{2024}+1}+\dfrac{2023}{2024^{2024}+1}=1+\dfrac{2023}{2024^{2024}+1}\)
\(B=\dfrac{2024^{2022}+1}{2024^{2023}+1}\)
\(2024B=\dfrac{2024^{2023}+2024}{2024^{2023}+1}=\dfrac{\left(2024^{2023}+1\right)+2023}{2024^{2023}+1}=\dfrac{2024^{2023}+1}{2024^{2023}+1}+\dfrac{2023}{2024^{2023}+1}=1+\dfrac{2023}{2024^{2023}+1}\)
Vì \(2024>2023=>2024^{2024}>2024^{2023}\)
\(=>2024^{2024}+1>2024^{2023}+1\)
\(=>\dfrac{2023}{2024^{2023}+1}>\dfrac{2023}{2024^{2024}+1}\)
\(=>A< B\)
\(#PaooNqoccc\)
24 tháng 7 2023
a: \(B=\dfrac{154}{155+156}+\dfrac{155}{155+156}\)
\(\dfrac{154}{155}>\dfrac{154}{155+156}\)
\(\dfrac{155}{156}>\dfrac{155}{155+156}\)
=>154/155+155/156>(154+155)/(155+156)
=>A>B
b: \(C=\dfrac{2021+2022+2023}{2022+2023+2024}=\dfrac{2021}{6069}+\dfrac{2022}{6069}+\dfrac{2023}{6069}\)
2021/2022>2021/6069
2022/2023>2022/2069
2023/2024>2023/6069
=>D>C
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
1 tháng 11
A = \(\dfrac{1}{2021.2022}\) + \(\dfrac{1}{2022.2023}\) + \(\dfrac{1}{2023.2024}\) + \(\dfrac{1}{2024.2025}\) - \(\dfrac{4}{2021.2025}\)
A = \(\dfrac{1}{2021}\) - \(\dfrac{1}{2022}\) + \(\dfrac{1}{2022}\) - \(\dfrac{1}{2023}\) + \(\dfrac{1}{2023}\) - \(\dfrac{1}{2024}\) + \(\dfrac{1}{2024}\) - \(\dfrac{1}{2025}\) - \(\dfrac{1}{2021}\) + \(\dfrac{1}{2025}\)
A = (\(\dfrac{1}{2021}\) - \(\dfrac{1}{2021}\)) + (\(\dfrac{1}{2022}\) - \(\dfrac{1}{2022}\)) + (\(\dfrac{1}{2023}\) - \(\dfrac{1}{2023}\)) + (\(\dfrac{1}{2024}\) - \(\dfrac{1}{2024}\)) + (\(\dfrac{1}{2025}\) - \(\dfrac{1}{2025}\))
A = 0 + 0 +0 + 0+ ... + 0
A = 0
DT
3
TH
Thầy Hùng Olm
Manager
VIP
2 tháng 5 2023
B = \(1-\dfrac{1}{2025}\) \(A=1-\dfrac{1}{2024}\)
Vì \(\dfrac{1}{2025}< \dfrac{1}{2024}\)
Nên B>A
NA
Nguyễn An Ninh
VIP
2 tháng 5 2023
Ta có :
\(\dfrac{2023}{2024}\)=\(\dfrac{2024-1}{2024}\)=\(\dfrac{2024}{2024}\)-\(\dfrac{1}{2024}\)=1-\(\dfrac{1}{2024}\)
\(\dfrac{2024}{2025}\)=\(\dfrac{2025-1}{2025}\)=\(\dfrac{2025}{2025}\)-\(\dfrac{1}{2025}\)=1=\(\dfrac{1}{2025}\)
Ta thấy: \(\dfrac{1}{2024}\) lớn hơn \(\dfrac{1}{2025}\)
Nên : \(\dfrac{2023}{2024}\) lớn hơn \(\dfrac{2024}{2025}\)
⇒A lớn hơn B
HA
1
BD
17 tháng 6 2023
\(A=2023\times2024\\ =\left(2022+1\right)\times2024\\ =2022\times2024+2024\\ B=2022\times2025\\ =2022\times\left(2024+1\right)\\ =2022\times2024+2022\)
Vì 2022 x 2024 = 2022 x 2024
=> 2024 > 2022
=> A> B
Cách 2
A= 2023 x 2024 = 4094552
B = 2022 x 2025 =4094550
Vì 4094552 > 4094550 = > A> B
NK
1
26 tháng 6 2023
Ta có : \(N=2022.2024\)
\(N=\left(2023-1\right)\left(2023+1\right)\)
\(N=2023^2+2023-2023-1\)
\(N=2023^2-1\)
Mà : \(M=2023.2023=2023^2\)
\(\Rightarrow M>N\)
HD
So sánh
A = \(\dfrac{2022^{2023}+1}{2022^{2024}+1}\) và B = \(\dfrac{2022^{2022}+1}{2022^{2023}+1}\)
2
4 tháng 7 2023
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.
4 tháng 7 2023
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
LT
1
11 tháng 9 2023
Ta có :
\(\dfrac{10^{2023}}{10^{2024}}=\dfrac{10^{2022}}{10^{2023}}\)
mà \(\dfrac{10^{2023}}{10^{2024}}>\dfrac{10^{2023}-3}{10^{2024}-3}\)
\(\dfrac{10^{2022}}{10^{2023}}< \dfrac{10^{2022}+1}{10^{2023}+1}\)
\(\Rightarrow\dfrac{10^{2023}-3}{10^{2024}-3}< \dfrac{10^{2022}+1}{10^{2023}+1}\)
Ta có:
\(A=\dfrac{2022}{2023}+\dfrac{2023}{2024}+\dfrac{2024}{2025}\\ =\left(1-\dfrac{1}{2023}\right)+\left(1-\dfrac{1}{2024}\right)+\left(1-\dfrac{1}{2025}\right)\\ =1-\dfrac{1}{2023}+1-\dfrac{1}{2024}+1-\dfrac{1}{2025}\\=\left(1+1+\right)-\left(\dfrac{1}{2023}+\dfrac{1}{2024}+\dfrac{1}{2025}\right)\\ =3-\left(\dfrac{1}{2023}+\dfrac{1}{2024}+\dfrac{1}{2025}\right)< 3\)
A = \(\dfrac{2022}{2023}\) + \(\dfrac{2023}{2024}\) + \(\dfrac{2024}{2025}\) Vì \(\dfrac{2022}{2023}\) < 1; \(\dfrac{2023}{2024}\) < 1; \(\dfrac{2024}{2025}\) < 1
Vậy A = \(\dfrac{2022}{2023}+\dfrac{2023}{2024}+\dfrac{2024}{2025}\) < 1 + 1 + 1 = 2 + 1 = 3
Vậy A < 3