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B = \(1-\dfrac{1}{2025}\) \(A=1-\dfrac{1}{2024}\)
Vì \(\dfrac{1}{2025}< \dfrac{1}{2024}\)
Nên B>A
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
\(A=\dfrac{2024^{2023}+1}{2024^{2024}+1}\)
\(2024A=\dfrac{2024^{2024}+2024}{2024^{2024}+1}=\dfrac{\left(2024^{2024}+1\right)+2023}{2024^{2024}+1}=\dfrac{2024^{2024}+1}{2024^{2024}+1}+\dfrac{2023}{2024^{2024}+1}=1+\dfrac{2023}{2024^{2024}+1}\)
\(B=\dfrac{2024^{2022}+1}{2024^{2023}+1}\)
\(2024B=\dfrac{2024^{2023}+2024}{2024^{2023}+1}=\dfrac{\left(2024^{2023}+1\right)+2023}{2024^{2023}+1}=\dfrac{2024^{2023}+1}{2024^{2023}+1}+\dfrac{2023}{2024^{2023}+1}=1+\dfrac{2023}{2024^{2023}+1}\)
Vì \(2024>2023=>2024^{2024}>2024^{2023}\)
\(=>2024^{2024}+1>2024^{2023}+1\)
\(=>\dfrac{2023}{2024^{2023}+1}>\dfrac{2023}{2024^{2024}+1}\)
\(=>A< B\)
\(#PaooNqoccc\)
\(A=2023\times2024\\ =\left(2022+1\right)\times2024\\ =2022\times2024+2024\\ B=2022\times2025\\ =2022\times\left(2024+1\right)\\ =2022\times2024+2022\)
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
b) \(M=\dfrac{10^{2023}+1}{10^{2024}+1}< 1\) ( Vì tử < mẫu )
Ta có: \(M=\dfrac{10^{2023}+1}{10^{2024}+1}< \dfrac{10^{2023}+1+9}{10^{2024}+1+9}=\dfrac{10^{2023}+10}{10^{2024}+10}=\dfrac{10.\left(10^{2022}+1\right)}{10.\left(10^{2023}+1\right)}=\dfrac{10^{2022}+1}{10^{2023}+1}=N\)
Vì \(\dfrac{10^{2023}+1}{10^{2024}+1}< \dfrac{10^{2022}+1}{10^{2023}+1}\) nên \(M< N\)
\(C=\dfrac{2^{2024}-3}{2^{2023}-1}=\dfrac{2.2^{2023}-2-1}{2^{2023}-1}=\dfrac{2\left(2^{2023}-1\right)-1}{2^{2023}-1}=2-\dfrac{1}{2^{2023}-1}\)
\(D=\dfrac{2^{2023}-3}{2^{2022}-1}=\dfrac{2.2^{2022}-2-1}{2^{2022}-1}=\dfrac{2\left(2^{2022}-1\right)-1}{2^{2022}-1}=2-\dfrac{1}{2^{2022}-1}\)
Ta có
\(2^{2023}>2^{2022}\Rightarrow2^{2023}-1>2^{2022}-1\)
\(\Rightarrow\dfrac{1}{2^{2023}-1}< \dfrac{1}{2^{2022}-1}\Rightarrow2-\dfrac{1}{2^{2023}-1}>2-\dfrac{1}{2^{2022}-1}\)
\(\Rightarrow C>D\)
a) \(2023^{2024}\) và \(2023^{2023}\)
vì 2024 > 2023 nên 20232024 > 20232023
Vậy 20232024 > 20232023
b) \(17^{2024}\) và \(18^{2024}\)
vì 17 < 18 nên 172024 < 18 2024
Vậy 172024 < 182024
Lời giải:
$-A=1\frac{2023}{2024}=1+\frac{2023}{2024}=1+1-\frac{1}{2024}$
$<1+1-\frac{1}{2025}=1+\frac{2024}{2025}=1\frac{2024}{2025}=-B$
$\Rightarrow A> B$