Các câu hỏi tương tự
1)
Cho \(\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{200}\)
Chứng minh: \(A>\frac{9}{10}\)
2)
Cho \(B=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)
Chứng minh: \(B>\frac{7}{12}\)
Chứng minh:
A = \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+.....+\frac{1}{199}+\frac{1}{200}>\frac{7}{12}\)
Chứng minh rằng: \(C=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}>\frac{7}{12}\)
cho A=\(\frac{1}{101}\)+\(\frac{1}{102}\)+\(\frac{1}{103}\)+...+\(\frac{1}{200}\)
chứng minh rằng : a, A>\(\frac{7}{12}\)
b, A>\(\frac{5}{8}\)
Chứng minh rằng:
a) \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)=\(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)
b) \(\frac{51}{2}+\frac{52}{2}+...+\frac{100}{2}=1.3.5...99\)
Chứng minh: \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}>\frac{7}{12}\)
Chứng minh: \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}<\frac{5}{8}\)
\(Cho A=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+\frac{1}{104}+.....+\frac{1}{200}\). Chứng tỏ: \(A>\frac{7}{12}\)
CHO
S=\(\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{199}+\frac{1}{200}\)
CHỨNG MINH RẰNG S>\(\frac{9}{10}\)