Các câu hỏi tương tự
CTR: \(\frac{1}{5}<\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-...+\frac{1}{98}-\frac{1}{99}<\frac{2}{5}\)
Chứng minh rằng
a) \(\frac{1}{5}<\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}<\frac{2}{5}\)
b) \(\frac{1}{15}<\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}<\frac{1}{10}\)
1. Tính tích N=\(\frac{1}{2}.\frac{-2}{3}.\frac{3}{4}.\frac{-4}{5}.\frac{5}{6}.\frac{-6}{7}.......\frac{-98}{99}.\frac{99}{100}\)
CMR: \(\frac{1}{5}<\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-...+\frac{1}{98}-\frac{1}{99}<\frac{2}{5}\)
bn nao nhanh mik tick nha
CMR
\(\frac{1}{5}< \frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}< \frac{2}{5}\)\(\frac{2}{5}\)
Cho \(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{99}{100}\)
CTR: \(\frac{1}{15}< A< \frac{1}{10}\)
16 tháng 2 2020 lúc 10:02
Tính : \(K=\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}\right).3^5+\left(\frac{1}{3^5}+\frac{1}{3^6}+\frac{1}{3^7}+\frac{1}{3^8}.3^9\right)+...+\left(\frac{1}{3^{97}}+\frac{1}{3^{98}}+\frac{1}{3^{99}}+\frac{1}{3^{100}}\right).3^{101}\)
CM B=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{98}-\frac{1}{99}>\frac{1}{5}\)
Chứng minh rằng :a, frac{1}{3}+frac{1}{31}+frac{1}{35}+frac{1}{37}+frac{1}{47}+frac{1}{53}+frac{1}{61}frac{1}{12}b, frac{1}{6}frac{1}{^{5^2}}+frac{1}{6^2}+frac{1}{7^2}+...+frac{1}{100^2}frac{1}{4}c, frac{1}{5}frac{1}{2}-frac{1}{3}+frac{1}{4}-frac{1}{5}+...+frac{1}{98}-frac{1}{99}frac{2}{5}d, 1frac{1}{1!}+frac{1}{2!}+frac{1}{3!}+...+frac{1}{100!}2
Đọc tiếp
Chứng minh rằng :
a, \(\frac{1}{3}+\frac{1}{31}+\frac{1}{35}+\frac{1}{37}+\frac{1}{47}+\frac{1}{53}+\frac{1}{61}<\frac{1}{12}\)
b, \(\frac{1}{6}<\frac{1}{^{5^2}}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}<\frac{1}{4}\)
c, \(\frac{1}{5}<\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}<\frac{2}{5}\)
d, \(1<\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}<2\)