Ta có: \(A=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{299}+\frac{1}{300}>\frac{1}{300}.200=\frac{2}{3}\Rightarrow A>\frac{2}{3}\Rightarrowđpcm\)

\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}>\frac{2}{3}\)

Biểu thức có 200 số hạng

Ta có: \(\frac{1}{101}>\frac{1}{300};\frac{1}{102}>\frac{1}{300};...;\frac{1}{299}>\frac{1}{300};\frac{1}{300}=\frac{1}{300}\)

\(\Rightarrow\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}>\frac{1}{300}+\frac{1}{300}+...+\frac{1}{300}=\frac{200}{300}=\frac{2}{3}\)

Vậy....

Ta có : \(\frac{1}{101}>\frac{1}{300}\)

            \(\frac{1}{102}>\frac{1}{300}\)

              ..................

              \(\frac{1}{300}=\frac{1}{300}\)

Do đó \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{299}+\frac{1}{300}>\frac{1}{300}+\frac{1}{300}+...+\frac{1}{300}\)

Hay     \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}>200\cdot\frac{1}{300}=\frac{2}{3}\Rightarrowđpcm\)

- Tham khảo ở đây đi : Câu hỏi của Nguyễn Thị Bích Phương - Toán lớp 6 | Học trực tuyến

Đặt A=\(\dfrac{1}{101}\)+\(\dfrac{1}{102}\)+\(\dfrac{1}{103}\)+...+\(\dfrac{1}{300}\)

Vì \(\dfrac{1}{101}\)>\(\dfrac{1}{102}\)>\(\dfrac{1}{103}\)>...>\(\dfrac{1}{300}\)

=>(\(\dfrac{1}{101}\)+\(\dfrac{1}{102}\)+\(\dfrac{1}{103}\)+...+\(\dfrac{1}{200}\))+(\(\dfrac{1}{201}\)+\(\dfrac{1}{202}\)+\(\dfrac{1}{203}\)+...+\(\dfrac{1}{300}\)) > (\(\dfrac{1}{200}\)+\(\dfrac{1}{200}\)+\(\dfrac{1}{200}\)+...+\(\dfrac{1}{200}\))+(\(\dfrac{1}{300}\)+\(\dfrac{1}{300}\)+\(\dfrac{1}{300}\)+...+\(\dfrac{1}{300}\)) =>\(\dfrac{1}{101}\)+\(\dfrac{1}{102}\)+\(\dfrac{1}{103}\)+...+\(\dfrac{1}{300}\) > \(\dfrac{1}{200}\).100 +\(\dfrac{1}{300}\) .100

=> A > \(\dfrac{1}{2}+\dfrac{1}{3}\)

=> A > \(\dfrac{5}{6}\) Mà \(\dfrac{5}{6}\)>\(\dfrac{2}{3}\)=> A > \(\dfrac{2}{3}\) Vậy \(\dfrac{1}{101}\)+\(\dfrac{1}{102}\)+\(\dfrac{1}{103}\)+...+\(\dfrac{1}{300}\) >\(\dfrac{2}{3}\)

\(A=\dfrac{1}{1.300}+\dfrac{1}{2.301}+...+\dfrac{1}{101.400}\)

\(\Rightarrow299A=\dfrac{299}{1.300}+\dfrac{299}{2.301}+...+\dfrac{299}{101.400}=1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+...+\dfrac{1}{101}-\dfrac{1}{400}=M\)

\(\Rightarrow A=\dfrac{M}{299}\left(1\right)\)

Ta lại có:

\(B=\dfrac{1}{1.102}+\dfrac{1}{2.103}+...+\dfrac{1}{298.399}+\dfrac{1}{299.400}\)

\(\Rightarrow101B=\dfrac{101}{1.102}+\dfrac{101}{2.103}+...+\dfrac{101}{399.400}=1-\dfrac{1}{102}+\dfrac{1}{2}-\dfrac{1}{103}+...+\dfrac{1}{399}-\dfrac{1}{400}=1-\dfrac{1}{300}+\dfrac{1}{2}-\dfrac{1}{301}+...+\dfrac{1}{101}-\dfrac{1}{400}=M\)

\(\Rightarrow B=\dfrac{M}{101}\left(2\right)\)

Từ \(\left(1\right),\left(2\right)\Rightarrow\dfrac{A}{B}=\dfrac{M}{299}:\dfrac{M}{101}=\dfrac{101}{299}\)