Đặt C = \(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{2015.2018}\)
\(\Rightarrow3C=\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{2015.2018}\)
\(\Rightarrow3C=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{2015}-\frac{1}{2018}\)
\(\Rightarrow3C=\frac{1}{2}-\frac{1}{2018}=\frac{504}{1009}\)
\(\Rightarrow C=\frac{504}{1009}:3=\frac{168}{1009}\)
Vậy \(C=\frac{168}{1009}\)
Tính \(S=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{101.104}\)
Tính nhanh :
B = \(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}\)
Tìm ybiết: y\(\times\)2.5+\(\frac{1}{4}\)=(1+\(\frac{1}{2}\)):(1+\(\frac{1}{3}\)):(1+\(\frac{1}{4}\)):.....:(1+\(\frac{1}{99}\))
A= 3/2.5+3/5.8+3/8.11+3/11.14
\(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{14}+\frac{1}{15}+\frac{1}{18}+\frac{1}{22}+\frac{1}{24}+\frac{1}{28}+\frac{1}{33}\)
Chứng minh : \(\frac{1}{1}-\frac{1}{1}+\frac{1}{2}-\frac{1}{3}+\frac{1}{5}-\frac{1}{8}+\frac{1}{13}-\frac{1}{21}+\frac{1}{34}-\frac{1}{55}+...< \frac{3}{10}\).
\(\frac{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}}{\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}}=?\)
\(\frac{\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{101}\right)-\left(\frac{1}{300}+\frac{1}{301}+\frac{1}{302}+...+\frac{1}{400}\right)}{\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{299}\right)-\left(\frac{1}{102}+\frac{1}{103}+\frac{1}{104}+...+\frac{1}{400}\right)}\)
G mk với, mk cần gấp lắm. Ai giải được mk k cho
\(\frac{\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}\right)\div\left(\frac{1}{6}+\frac{1}{10}-\frac{1}{15}\right)}{\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)\div\left(\frac{1}{4}-\frac{1}{6}\right)}\)
