`A = 2024/(1*2) + 2024/(2*3) + ... + 2024/(2023*2024)`
`=> A =2024 * ( 1/(1*2) + 1/(2*3) +... +1/(2023*2024) )`
`=> A =2024 *(1 -1/2 + 1/2 -1/3 +... +1/2023 -1/2024)`
`=> A =2024 * (1 -1/2024)`
`=> A =2024 * 2023/2024`
`=> A=2023`
Vậy `A= 2023`
\(x-\dfrac{1}{2}-\dfrac{1}{2.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}-...-\dfrac{1}{2021.2022}-\dfrac{1}{2022.2023}=\dfrac{-2024}{2023}\)
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!
\(\dfrac{2022}{2023}\)+\(\dfrac{2023}{2024}\)+\(\dfrac{2024}{2022}\)
so sánh
B=\(\dfrac{2024}{2025}\)
A=\(\dfrac{2023}{2024}\)
A = \(\dfrac{1}{3}\)-\(\dfrac{2}{^{ }3^2}\)+\(\dfrac{3}{3^3}\)-\(\dfrac{4}{3^4}\)+...+\(\dfrac{2023}{3^{2023}}\)-\(\dfrac{2024}{3^{2024}}\) so sánh A với \(\dfrac{3}{16}\)
Chứng minh M thuộc tập số nguyên, biết M = \(\dfrac{1}{3}\) + \(\dfrac{2}{3^2}\) + \(\dfrac{3}{3^3}\) + ... + \(\dfrac{2024}{3^{2024}}\)
\(\dfrac{2^{2023}+3^{2023}}{2^{2024}+3^{2024}}\) chứng minh phấn số đó tối giản
\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{2024}\)
1: \(\dfrac{2}{3}\): \(\dfrac{3}{4}\) :\(\dfrac{4}{5}\) :.....:\(\dfrac{2024}{2025}\)
Tính nhanh: \(\dfrac{7}{1`2}.\dfrac{2024}{2023}-\dfrac{7}{2023}.\dfrac{1}{2}\)
