So sánh :
a, 2023^2024 và 2023^2023
b,17^2024 và 18^2024
giúp tớ với !!! tớ gấp lắm rồi!
Giúp mình với!!!
So sánh A=\(\dfrac{2024^{2023}+1}{2024^{2024}+1}\) và B=\(\dfrac{2024^{2022}+1}{2024^{2023}+1}\)
Cám ơn các bạn!
\(A=\dfrac{2024^{2023}+1}{2024^{2024}+1}\)
\(2024A=\dfrac{2024^{2024}+2024}{2024^{2024}+1}=\dfrac{\left(2024^{2024}+1\right)+2023}{2024^{2024}+1}=\dfrac{2024^{2024}+1}{2024^{2024}+1}+\dfrac{2023}{2024^{2024}+1}=1+\dfrac{2023}{2024^{2024}+1}\)
\(B=\dfrac{2024^{2022}+1}{2024^{2023}+1}\)
\(2024B=\dfrac{2024^{2023}+2024}{2024^{2023}+1}=\dfrac{\left(2024^{2023}+1\right)+2023}{2024^{2023}+1}=\dfrac{2024^{2023}+1}{2024^{2023}+1}+\dfrac{2023}{2024^{2023}+1}=1+\dfrac{2023}{2024^{2023}+1}\)
Vì \(2024>2023=>2024^{2024}>2024^{2023}\)
\(=>2024^{2024}+1>2024^{2023}+1\)
\(=>\dfrac{2023}{2024^{2023}+1}>\dfrac{2023}{2024^{2024}+1}\)
\(=>A< B\)
\(#PaooNqoccc\)
bài 7 so sánh A và B
A=2022/2023 + 2023/2024 B=2022+2023/2023+2024
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
So sánh các cặp số sau:
\dfrac{ -2024 }{ 2023 } và \dfrac{ -2023 }{ 2024 }
-2024/2023<-1
-1<-2023/2024
=>-2024/2023<-2023/2024
so sánh
B=\(\dfrac{2024}{2025}\)
A=\(\dfrac{2023}{2024}\)
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
So sánh A=10^2024+1/10^2023+1 và B=10^2023+1=10^2022+1
\(A=\dfrac{10^{2024}+1}{10^{2023}+1}=\dfrac{10\left(10^{2023}+1\right)}{10^{2023}+1}-\dfrac{9}{10^{2023}+1}=1-\dfrac{9}{10^{2023}+1}\)
\(B=\dfrac{10^{2023}+1}{10^{2022}+1}=\dfrac{10\left(10^{2022}+1\right)}{10^{2022}+1}-\dfrac{9}{10^{2022}+1}=1-\dfrac{9}{10^{2022}+1}\)
Vì \(\dfrac{9}{10^{2023}+1}< \dfrac{9}{10^{2022}+1}\)
\(\Rightarrow A>B\)
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
hãy so sánh 2023^20 + 2023^19 và 2024^20 đó làm dược bày tui với