Các câu hỏi tương tự
chứng minh rằng : 1.3.5.7....197.199 = \(\frac{101}{2}.\frac{102}{2}.\frac{103}{2}....\frac{200}{2}\)
Chứng tỏ rằng :1.5.7...197.199=\(\frac{101}{2}.\frac{102}{2}.\frac{103}{2}...\frac{200}{2}\)
Chứng minh rằng:
1.3.5.7.9.....197.199=\(\frac{101}{2}+\frac{102}{2}+\frac{103}{2}+...+\frac{200}{2}\)
Tính: \(\frac{101}{2}.\frac{102}{6}.\frac{103}{10}...\frac{200}{398}\)
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{200}=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{200}=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)1/200.
giải hộ mình nha cám ơn
CMR:
\(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{299}+\frac{1}{300}>\frac{2}{3}\)\(\frac{2}{3}\)
CHỨNG MINH
a, 1.3.5.7.....197.199<\(\frac{101.102.102.....200}{1+2+2^2+2^3+...+2^{99}}\)
b, \(\frac{-1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{2499}{2500}>\frac{-1}{\left(-7\right)^2}\)
Chứng tỏ rằng \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+....+\frac{1}{200}>\frac{1}{2}\)
Chi tiết rõ ràng nha