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

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

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)

S=\(\left(\frac{1}{101}+\frac{1}{102}+....+\frac{1}{110}\right)\)  + \(\left(\frac{1}{111}+...+\frac{1}{120}\right)\) + \(\left(\frac{1}{121}+...+\frac{1}{130}\right)\)

> \(\frac{1}{110}.10+\frac{1}{120}.10+\frac{1}{130.10}=\)\(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}\)> \(\frac{1}{12}+\frac{2}{12}=\frac{1}{4}\) ( TA CÓ:\(\frac{1}{11}+\frac{1}{13}>\frac{2}{12}\))

\(\Rightarrow S>\frac{1}{4}\)(1)

+)S=\(\left(\frac{1}{101}+\frac{1}{130}\right)+\left(\frac{1}{102}+\frac{1}{129}\right)+...+\) \(\left(\frac{1}{115}+\frac{1}{116}\right)\) (CÓ 15 Cặp)

=\(\left(\frac{231}{101.130}\right)+\left(\frac{231}{102.129}\right)+...+\)\(\left(\frac{231}{115.116}\right)\)=\(231.\left(\frac{1}{101.130}+\frac{1}{102.129}+...+\frac{1}{115.116}\right)\)

ta xét: tích 101.130 có giá trị nhỏ nhất,nên :

xét 101.129=(101+1).(101-1)=101.130-101+130-1=101.130+28>101.130

tương tự các cặp còn lại, vậy ta có:\(\frac{1}{101.130}+\frac{1}{120.129}+...+\frac{1}{115.116}< \frac{1}{101.130}.15\)

\(\Rightarrow S< 231.\frac{1}{101.130}.15=\frac{693}{2626}< \frac{91}{330}\left(2\right)\)

từ (1)và(2) \(\Rightarrow\)điều phải chứng  minh

THANKS

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

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

Ta có:

\(A=\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}+\dfrac{1}{201}+\dfrac{1}{202}+...+\dfrac{1}{300}\)

Do: \(\dfrac{1}{101}< \dfrac{1}{100}\); \(\dfrac{1}{102}< \dfrac{1}{100}\); ...; \(\dfrac{1}{200}< \dfrac{1}{100}\)

\(\Rightarrow\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}< \dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}\)

\(\Rightarrow\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}< \dfrac{100}{100}=1\) (1)

Lại có:

\(\dfrac{1}{201}< \dfrac{1}{200}\) ; \(\dfrac{1}{202}< \dfrac{1}{200}\) ;...;\(\dfrac{1}{300}< \dfrac{1}{200}\)

\(\Rightarrow\dfrac{1}{201}+\dfrac{1}{202}+...+\dfrac{1}{300}< \dfrac{1}{200}+\dfrac{1}{200}+...+\dfrac{1}{200}\)

\(\Rightarrow\dfrac{1}{201}+\dfrac{1}{202}+...+\dfrac{1}{300}< \dfrac{100}{200}=\dfrac{1}{2}\) (2)

Từ (1);(2) \(\Rightarrow\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{300}< 1+\dfrac{1}{2}\)

\(\Rightarrow A< \dfrac{3}{2}\)

\(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}\)( có 200 số )

Ta có

\(\frac{1}{101}>\frac{1}{300}\); \(\frac{1}{102}>\frac{1}{300}\); ...;\(\frac{1}{299}>\frac{1}{300}\)

=> \(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}\)> \(\frac{1}{300}+\frac{1}{300}+...+\frac{1}{300}+\frac{1}{300}\)

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

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

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

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

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