Các câu hỏi tương tự
Tính E=\(\frac{\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+....+\frac{1}{100}}{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{99.100}}\)
Tính:
\(P=\frac{\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}}{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}}\)
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}.\)
CM. \(\frac{5}{6}< A< \frac{7}{12}\)
cho A = \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+.......+\frac{1}{99.100}\)
CMR: \(̃̃̃̃\frac{7}{12}< A< \frac{5}{6}\)
Tính \(E=\frac{\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+....+\frac{1}{100}}{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+..+\frac{1}{99.100}}\)
Chứng minh:
\(\frac{7}{12}< \frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}< \frac{5}{6}\)
tính A-B biết
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{97.98}+\frac{1}{99.100}\)
\(B=\frac{1}{50}+\frac{1}{51}+...+\frac{1}{99}+\frac{1}{100}\)
\(P=\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right):\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\right)\)
tinh p
CMR: \(\frac{1}{1.2}+\frac{1}{3.4}\)\(+\frac{1}{5.6}+...+\frac{1}{99.100}=\frac{1}{51}+\frac{1}{52}\)\(+\frac{1}{53}+...+\frac{1}{100}\)