Các câu hỏi tương tự
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{200}=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)
Chứng tỏ rằng :1.5.7...197.199=\(\frac{101}{2}.\frac{102}{2}.\frac{103}{2}...\frac{200}{2}\)
CMR: 1.3.5.7......197.199=\(\frac{101}{2}+\frac{102}{2}+\frac{103}{2}+........+\frac{200}{2}\)
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{999}-\frac{1}{200}=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)1/200.
giải hộ mình nha cám ơn
chứng minh rằng : 1.3.5.7....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:
.\(\frac{\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}}{\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{199.200}}\)
C/m:\(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+....+\frac{1}{200}>\frac{7}{12}\)
Chứng minh: \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}<\frac{5}{8}\)