1256+2024=
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
10 tháng 9 2023
\(S=C^0_{2024}+\dfrac{1}{2}C^2_{2024}+\dfrac{1}{3}C^4_{2024}+\dfrac{1}{4}C^6_{2024}+...+\dfrac{1}{1013}C^{2024}_{2024}\)
Ta có :
\(\dfrac{1}{k+1}C^{2k-1}_n=\dfrac{1}{k+1}.\dfrac{n!}{\left(2k-1\right)!\left(n-2k+1\right)!}\)
\(=\dfrac{1}{n+1}.\dfrac{\left(n+1\right)!}{2k!\left[\left(n+1\right)-2k\right]!}\)
\(=\dfrac{1}{n+1}C^{2k}_{n+1}\)
\(\Rightarrow S_n=\dfrac{1}{n+1}\Sigma^{2k}_{k=0}C^{2k}_{n+1}=\dfrac{1}{n+1}\left(\Sigma^{2k}_{k=0}C^{2k-1}_{n+1}-C^0_{n+1}\right)=\dfrac{2^{2n-1}-1}{n+1}\)
\(\Rightarrow S=\dfrac{2^{2025}-1}{1013}\)
M
10 tháng 9 2023
S = C₀₂₀₂₄ + 12.C₂₀₂₄ + 13.C₂₀₂₄ + 14.C₂₀₂₄ + ... + 11013.C₂₀₂₄
= (C₀₂₀₂₄ + C₂₀₂₄ + C₂₀₂₄ + C₂₀₂₄ + ... + C₂₀₂₄) + (C₂₀₂₄ + C₂₀₂₄ + C₂₀₂₄ + ... + C₂₀₂₄) + ... + (C₂₀₂₄)
= 11014.C₂₀₂₄
= 11014.
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 \(2024>2023=>2024^{2024}>2024^{2023}\)
\(=>2024^{2024}+1>2024^{2023}+1\)
\(=>\dfrac{2023}{2024^{2023}+1}>\dfrac{2023}{2024^{2024}+1}\)
\(=>A< B\)
\(#PaooNqoccc\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 1 2024
\(\sqrt{x^2+2024}=\sqrt{x^2+xy+yz+zx}=\sqrt{\left(x+y\right)\left(z+x\right)}\ge\sqrt{\left(\sqrt{xz}+\sqrt{xy}\right)^2}=\sqrt{xy}+\sqrt{xz}\)
Tương tự: \(\sqrt{y^2+2024}\ge\sqrt{xy}+\sqrt{yz}\)
\(\sqrt{z^2+2024}\ge\sqrt{xz}+\sqrt{yz}\)
Cộng vế:
\(P\ge\dfrac{2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)}{\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}=2\)
Dấu "=" xảy ra khi \(x=y=z=\dfrac{2024}{3}\)
=3280
3280