Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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 \(\dfrac{1}{101}\)+\(\dfrac{1}{102}\)+\(\dfrac{1}{103}\)+...+\(\dfrac{1}{300}\) >\(\dfrac{2}{3}\)
Chứng minh 1/101 + 1/102 + ... + 1/299 + 1/300 > 2/3
Ta có:
1/101>1/300
1/102>1/300
.....
1/299>1/300
=>VT>200.1/300=200/300=2/3(dpcm)
Chứng minh 1/101 + 1/102 + ... + 1/299 + 1/300 > 2/3
Ta có: 1/101> 1/300; 1/102> 1/300; .....; 1/300= 1/300
1/101 + 1/102 + ... + 1/299 + 1/300 > 1/300+ 1/300+ .........+1/300= 200/300= 2/3
Vậy 1/101 + 1/102 + ... + 1/299 + 1/300 > 2/3 (dpcm)
refer
https://hoc247.net/hoi-dap/toan-6/chung-minh-1-101-1-102-1-103-1-104-1-299-1-300-2-3-faq302038.html
\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}>\frac{2}{3}\)
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 \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}>200\cdot\frac{1}{300}=\frac{2}{3}\Rightarrowđpcm\)