28 tháng 2 2016 lúc 18:03
Các câu hỏi tương tự
\(\frac{\frac{4000}{1}+\frac{3999}{2}+\frac{3998}{3}+...+\frac{1}{4000}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{4001}}\)=?
\(y=\frac{\frac{4000}{1}+\frac{3999}{2}+\frac{3998}{3}+...+\frac{1}{4000}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{4001}}=?\)
Rút gọn: 4000/1+3999/2+3998/3+...+1/4000 / 1/2+1/3+1/4+...+1/4001
Tìm x, biết:
\(\left(\frac{1999}{2}+\frac{1998}{3}+\frac{1997}{4}+.......+\frac{1}{2000}+4000\right)x=1+\frac{1}{2}+\frac{1}{3}\)\(\frac{1}{3}\)
Rút gọn:
\(\frac{2016-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-...-\frac{1}{2017}}{\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{2015}{2016}}\)
Rút gọn:
\(\frac{\frac{1}{2020}+\frac{2}{2019}+\frac{3}{2018}+...+\frac{2019}{2}+\frac{2020}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}\)
Rút gọn:
\(B=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+.....+\frac{1}{3^{99}}\)
Rút gọn các biểu thức sau:\(\frac{\frac{1}{2}-\frac{1}{3}+\frac{2}{5}+\frac{1}{8}}{\frac{1}{6}+\frac{3}{20}-\frac{1}{3}+\frac{3}{4}}\)
Rút gọn
\(\frac{3-\frac{1}{5}+\frac{3}{20}}{2+\frac{1}{4}-\frac{3}{5}}\)
Tính
a/ \(\frac{3}{4}.\frac{8}{9}.\frac{15}{10}....\frac{9999}{1000}\)
b/ \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< 1\)