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T
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 \(2024>2023=>2024^{2024}>2024^{2023}\)
\(=>2024^{2024}+1>2024^{2023}+1\)
\(=>\dfrac{2023}{2024^{2023}+1}>\dfrac{2023}{2024^{2024}+1}\)
\(=>A< B\)
\(#PaooNqoccc\)
DN
3
22 tháng 8 2023
a) \(\left(x-2024\right)^{2023}=1\)
\(\Rightarrow\left(x-2024\right)^{2023}=1^{2023}\)
\(\Rightarrow x-2024=1\)
\(\Rightarrow x=2025\)
b) \(\left(2x-1\right)^5=32\)
\(\Rightarrow\left(2x-1\right)^5=2^5\)
\(\Rightarrow2x-1=2\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\dfrac{3}{2}\)
c) \(5< 2^x< 100\)
\(\Rightarrow4=2^2< 5< 2^x< 100< 128=2^7\)
\(\Rightarrow2< x< 7\)
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
NS
2
29 tháng 10 2023
1-2+3-4+5-6+7-8+...+2023-2024
\(=\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+\left(7-8\right)+....+\left(2023-2024\right)\)
\(=-1+\left(-1\right)+\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)
\(=-1.1012\)
\(=-1012\)
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