Mai anh ê tick cho mình đi

\(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{99}\right)\)

\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}....\dfrac{98}{99}=\dfrac{1.2.3...98}{2.3.4...99}=\dfrac{1}{99}\)

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 !!!

Tra lời:

Ta có:

1/101➢1/300+1/102➢1/300+1/103➢1/300+1/104➢1/300+.....+1/299➢1/300

=1/101+1/102+1/103+...1/299➢199/300

=1/101+1/102+1/103+...1/299+1/300➢199/300+1/300

=200/300=2/3.

Note: ➢ là dau lớn do nhe. Nho tick cho minh nha😊😉

\(\frac{1}{101}\)\(+\)\(\frac{1}{102}\)\(+\). . . . \(+\)\(\frac{1}{299}\)\(+\)\(\frac{1}{300}\)\(\ge\)\(\frac{2}{3}\)\(\ge\)\(\frac{1}{300}\)\(+\)\(\frac{1}{300}\)\(+\)\(\frac{1}{300}\)\(=\)\(\frac{200}{300}\)\(=\)\(\frac{2}{3}\)

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

\(\ge\)\(\frac{2}{3}\)

Ta có: \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{209}+\frac{1}{300}>\frac{1}{300}.200\)

\(\Rightarrow\frac{1}{101}+\frac{1}{102}+...+\frac{1}{209}+\frac{1}{300}>\frac{2}{3}\)

Vậy \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{209}+\frac{1}{300}>\frac{2}{3}\left(đpcm\right)\)

đpcm là j z

điều phải chứng minh nha

- 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 \(\dfrac{1}{101}\)+\(\dfrac{1}{102}\)+\(\dfrac{1}{103}\)+...+\(\dfrac{1}{300}\) >\(\dfrac{2}{3}\)