Ta có: A =\(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{199}+\frac{1}{200}\)
=\(\left(\frac{1}{101}+\frac{1}{102}...+\frac{1}{150}\right)+\left(\frac{1}{151}+\frac{1}{152}+...+\frac{1}{200}\right)\)<
<\(\frac{1}{150}.50+\frac{1}{200}.50=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
K CHO MIK VS NHÉ!
Chứng tỏ:
\(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{199}+\frac{1}{200}\) < 1
\(Cho A=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+\frac{1}{104}+.....+\frac{1}{200}\). Chứng tỏ: \(A>\frac{7}{12}\)
Chứng minh:
A = \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+.....+\frac{1}{199}+\frac{1}{200}>\frac{7}{12}\)
A=\(\frac{1}{101}+\frac{1}{102}+............+\frac{1}{199}+\frac{1}{200}\)
Chứng tỏ A<\(\frac{5}{6}\)
Chứng tỏ :
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
SO SÁNH:
A=\(\frac{1}{101}\)+\(\frac{1}{102}\)+\(\frac{1}{103}\)+\(\frac{1}{104}\)+.......+\(\frac{1}{199}\)+\(\frac{1}{200}\)VỚI \(\frac{1}{2}\)
\(\text{Chứng tỏ:}\)
\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{199}-\frac{1}{200}=\frac{1}{101}+\frac{1}{102}+.....+\frac{1}{200}\)
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 tỏ rằng:
\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{199}+\frac{1}{200}< 1\)
Giúp mình ý!!!Mình cho like!!
