\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=\dfrac{1}{2}\cdot\dfrac{4}{10}=\dfrac{2}{10}=\dfrac{1}{5}\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{10}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=\dfrac{1}{2}\cdot\dfrac{4}{10}=\dfrac{2}{10}=\dfrac{1}{5}\)
\(\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+...+\dfrac{1}{96.98}+\dfrac{1}{98.100}\)
\(l,\dfrac{\dfrac{3}{41}-\dfrac{12}{47}+\dfrac{27}{53}}{\dfrac{4}{41}-\dfrac{16}{47}+\dfrac{36}{53}}\)
\(m,\left(3-2\dfrac{1}{3}+\dfrac{1}{4}\right):\left(4-5\dfrac{1}{6}+2\dfrac{1}{4}\right)\)
\(n,F=\dfrac{4}{2.4}+\dfrac{4}{4.6}+\dfrac{4}{6.8}+...+\dfrac{4}{2008.2010}\)
\(p,F=\dfrac{1}{18}+\dfrac{1}{54}+\dfrac{1}{108}+...+\dfrac{1}{990}\)
Cho A= \(\dfrac{4}{2.4}\)+\(\dfrac{4}{4.6}\)+\(\dfrac{4}{6.8}\)+...+\(\dfrac{4}{46.48}\)+\(\dfrac{4}{48.50}\)
Chứng tỏ :
a, A = \(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{2022.2024}\) < \(\dfrac{1}{4}\)
b, B =\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{2013.2015}< \dfrac{1}{2}\)
c, C =\(\dfrac{1}{3^2}+\dfrac{1}{5^2}+\dfrac{1}{7^2}+...+\dfrac{1}{2013^2}< \dfrac{1}{4}\)
d, D =\(\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{2014^2}< \dfrac{1}{2}\)
1. TÍNH A) A=\(\dfrac{1.3}{2^2}\).\(\dfrac{2.4}{3^2}\)......\(\dfrac{8.10}{9^2}\)
Tìm x, biết:
\(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{\left(2x-2\right).2x}=\dfrac{11}{48}\) (x ϵ N , x ≥ 2)
Tìm x
\(\dfrac{1}{2.4}\) + \(\dfrac{1}{4.6}\) +...+ \(\dfrac{1}{\left(2x-2\right).2x}\) = \(\dfrac{1}{8}\) ( x ∈ N , x ≥ 2 )
có lời giải chi tiết
S=1/2.4+1/4.6+1/6.8+1/8.10 tính S