\(1:\dfrac{2}{3}:\dfrac{3}{4}:\dfrac{4}{5}:...:\dfrac{2024}{2025}\)
= \(1\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{2025}{2024}=\dfrac{2025}{2}\)
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 \(\dfrac{3}{16}\)
Đố ai giải được câu này ko dùng google
B = 1 - \(\dfrac{3}{4}\) + (\(\dfrac{3}{4}^2\)) - (\(\dfrac{3}{4}^3\)) + ... - (\(\dfrac{3}{4}^{2023}\)) + (\(\dfrac{3}{4}^{2024}\))
so sánh
B=\(\dfrac{2024}{2025}\)
A=\(\dfrac{2023}{2024}\)
Chứng minh M thuộc tập số nguyên, biết M = \(\dfrac{1}{3}\) + \(\dfrac{2}{3^2}\) + \(\dfrac{3}{3^3}\) + ... + \(\dfrac{2024}{3^{2024}}\)
Bài 2: So sánh các phân số sau mà không quy đồng mẫu số và tử số:
a)\(\dfrac{2025}{2022}\)và\(\dfrac{2024}{2021}\)
b)\(\dfrac{8^9+1}{8^{10}+1}\)và \(\dfrac{8^8+1}{8^9+1}\)
giúp mk với nhé
\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{2024}\)
a) \(\dfrac{2}{5}+\dfrac{4}{3}:\dfrac{-2}{3}\)
b) \(3\dfrac{4}{5}-\left(2\dfrac{1}{4}+1\dfrac{4}{5}\right)\)
c) \(\dfrac{-3}{5}.\dfrac{4}{7}+\dfrac{3}{7}.\dfrac{-3}{5}+\dfrac{3}{5}\)
d) 40% \(-1\dfrac{5}{7}:\dfrac{3}{7}+\left|\dfrac{-9}{5}\right|\)
1, \(\dfrac{3}{4}\). ( \(\dfrac{2}{5}\) - \(\dfrac{1}{15}\)) +\(\dfrac{3}{4}\)
2, \(\dfrac{4}{9}\). (\(\dfrac{-13}{3}\)) + \(\dfrac{4}{3}\). \(\dfrac{40}{9}\)
3, \(\dfrac{4}{9}\) - \(\dfrac{2}{3}\). ( \(\dfrac{4}{5}\)+\(\dfrac{1}{2}\) )
giúp mình nha cảm ơn
1, \(\dfrac{-5}{7}\) . \(\dfrac{2}{11}\) + \(\dfrac{-5}{7}\) . \(\dfrac{9}{11}\) + \(1\dfrac{5}{7}\)
2,\(-3\dfrac{4}{13}\) . \(15\dfrac{3}{41}\) + \(3\dfrac{4}{13}\) . \(2\dfrac{3}{41}\)
3, \(\dfrac{4}{5}\) . \(15\dfrac{1}{4}\) - \(\dfrac{4}{5}\) . \(15\dfrac{1}{3}\) + \(\dfrac{11}{30}\)
4,\(\dfrac{4}{20}\) + \(\dfrac{16}{42}\) + \(\dfrac{6}{15}\) - \(\dfrac{3}{5}\) + \(\dfrac{2}{21}\) - \(\dfrac{10}{21}\) + \(\dfrac{3}{10}\)
Giúp mik nha. Cảm ơn
