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\(\frac{1}{101}\)\(+\)\(\frac{1}{102}\)\(+\). . . . \(+\)\(\frac{1}{299}\)\(+\)\(\frac{1}{300}\)\(\ge\)\(\frac{2}{3}\)\(\ge\)\(\frac{1}{300}\)\(+\)\(\frac{1}{300}\)\(+\)\(\frac{1}{300}\)\(=\)\(\frac{200}{300}\)\(=\)\(\frac{2}{3}\)
do \(\frac{1}{101}\)..... \(\frac{1}{300}\)có 200 số
\(\Rightarrow\)\(\frac{1}{101}\)\(+\)\(\frac{1}{102}\)\(+\)..... \(+\)\(\frac{1}{299}\)\(+\)\(\frac{1}{300}\)\(\ge\)\(\frac{1}{300}\)\(\times\)200
\(\ge\)\(\frac{2}{3}\)
\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}>\frac{2}{3}\)
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 \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}>200\cdot\frac{1}{300}=\frac{2}{3}\Rightarrowđpcm\)
Đặt\(A=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}\)
Vì\(\frac{1}{101}>\frac{1}{102}>\frac{1}{103}>...>\frac{1}{300}\)
\(\Rightarrow\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)+\left(\frac{1}{201}+\frac{1}{202}+...+\frac{1}{300}\right)\)\(>\left(\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\right)+\left(\frac{1}{300}+\frac{1}{300}+...+\frac{1}{300}\right)\)(mỗi cái trong ngoặc là một trăm phân số)
\(\Rightarrow\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}>\left(\frac{1}{200}\right).100+\left(\frac{1}{300}\right).100\)
\(\Rightarrow A>\frac{1}{2}+\frac{1}{3}\)
\(\Rightarrow A>\frac{5}{6}\)
Mà 5/6>2/3=>A>2/3
\(\Rightarrow\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{300}\)
Đặt A = \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{300}\)
Vì \(\frac{1}{101}>\frac{1}{102}>\frac{1}{103}>...>\frac{1}{300}\)
\(\Rightarrow\left(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+....\frac{1}{200}\right)+\left(\frac{1}{201}+\frac{1}{202}+\frac{1}{103}+.....\frac{1}{300}\right)>\left(\frac{1}{200}+\frac{1}{200}+\frac{1}{200}\right)\)
Tự làm tiếp nhé !!!
Ta có: \(A=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{299}+\frac{1}{300}>\frac{1}{300}.200=\frac{2}{3}\Rightarrow A>\frac{2}{3}\Rightarrowđpcm\)
- 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 \(\dfrac{1}{101}\)+\(\dfrac{1}{102}\)+\(\dfrac{1}{103}\)+...+\(\dfrac{1}{300}\) >\(\dfrac{2}{3}\)