Đặt A=1/101+1/102+1/103+...+1/300
vì 1/101>1/102>1/103>...>1/300
=>(1/101+1/102+1/103+...+1/200)+(1/201+1/202+1/103+...+1/300) > (1/200+1/200+1/200+...+1/200)+(1/300+1/300+1/300+...+1/300) 
=>1/101+1/102+1/103+...+1/300 > (1/200).100 + (1/300).100
=> A > 1/2+1/3
=> A > 5/6 
Mà 5/6>2/3
=> A > 2/3
Vậy 1/101+1/102+1/103+...+1/300 >2/3

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

\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{299}+\frac{1}{300}=\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)+\left(\frac{1}{201}+\frac{1}{202}+...++\frac{1}{299}+\frac{1}{300}\right)\)

                                                           \(=\left(\frac{1}{200}.100\right)+\left(\frac{1}{300}.100\right)\)

                                                         \(=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}>\frac{4}{6}=\frac{2}{3}\)

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

Sai dấu

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😊😉

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

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

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

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

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