​Ta có: 151+152+...+175>175+175+...+175=2575=13​

176+177+...+1100>1100+1100+...+1100=25100=14

​=> S>13+14=712​ (1)

Ta có: 151+152+...+175<150+150+...+150=2550=12​

176+177+...+1100<175+175+...+175=2575=13​

=> S<12+13=56 ​(2)

Từ (1) và (2) => 712​ < S<56 ​( đpcm )

Ta có:

- 1/51 > 1/75, 1/52 > 1/75 ...

=> 1/51 + 1/52 + ... + 1/75 > 1/75 + ... 1/75 = 25/75 = 1/3

- 1/76 > 1/100, 1/77 > 1/100 ...

=> 1/76 + 1/77 + ... + 1/100 > 1/100 + ... + 1/100 = 25/100 = 1/4

Từ đó : S = ( 1/51 + ... + 1/75 ) + ( 1/76 + ... + 1/100 ) > 1/3 + 1/3 = 7/12 (1)

- 1/51 < 1/50, 1/52 < 1/50 ... 

=> 1/51 + 1/52 + ... + 1/75 < 1/50 + ... 1/50 = 25/50 = 1/2

- 1/76 < 1/75, 1/77 < 1/75...

=> 1/76 + 1/77 + ... + 1/100 < 1/75 + ... + 1/75 = 25/75 = 1/3

Từ đó : S = ( 1/51 + ... + 1/75 ) + ( 1/76 + ... + 1/100 ) < 1/2 + 1/3 = 5/6 (2)

từ (1) và (2) => 5/6 > S > 7/12

* Chúc bn học tốt !!!

Ta có: \(\frac{1}{51}>\frac{1}{75};\frac{1}{52}>\frac{1}{75};\ldots;\frac{1}{74}>\frac{1}{75};\frac{1}{75}=\frac{1}{75}\)

Do đó: \(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{75}>\frac{1}{75}+\frac{1}{75}+\cdots+\frac{1}{75}=\frac{25}{75}=\frac13\) (1)

Ta có: \(\frac{1}{76}>\frac{1}{100};\frac{1}{77}>\frac{1}{100};\ldots;\frac{1}{99}>\frac{1}{100};\frac{1}{100}=\frac{1}{100}\)

Do đó: \(\frac{1}{76}+\frac{1}{77}+\cdots+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+\cdots+\frac{1}{100}=\frac{25}{100}=\frac14\) (2)

Từ (1),(2) ta có: \(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{75}+\frac{1}{76}+\frac{1}{77}+\cdots+\frac{1}{100}>\frac13+\frac14\)

=>\(S>\frac13+\frac14=\frac{7}{12}\) (3)

Ta có: \(\frac{1}{51}<\frac{1}{50};\frac{1}{52}<\frac{1}{50};\ldots;\frac{1}{75}<\frac{1}{50}\)

Do đó: \(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{75}<\frac{1}{50}+\frac{1}{50}+\cdots+\frac{1}{50}=\frac{25}{50}=\frac12\) (4)

Ta có: \(\frac{1}{76}<\frac{1}{75};\frac{1}{77}<\frac{1}{75};\ldots;\frac{1}{100}<\frac{1}{75}\)

Do đó: \(\frac{1}{76}+\frac{1}{77}+\cdots+\frac{1}{100}<\frac{1}{75}+\frac{1}{75}+\cdots+\frac{1}{75}=\frac{25}{75}=\frac13\) (5)

Từ (4),(5) suy ra \(\frac{1}{51}+\frac{1}{52}+\cdots+\frac{1}{75}+\frac{1}{76}+\frac{1}{77}+\cdots+\frac{1}{100}<\frac12+\frac13\)

=>\(S<\frac56\) (6)

Từ (3),(6) suy ra 7/12<S<5/6

Đặt \(A=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)

Ta có: \(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{75}>\dfrac{1}{75}+\dfrac{1}{75}+...+\dfrac{1}{75}=\dfrac{25}{75}=\dfrac{1}{3}\)

\(\dfrac{1}{76}+\dfrac{1}{77}+...+\dfrac{1}{100}>\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{25}{100}=\dfrac{1}{4}\)

Do đó: \(A>\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}\)(1)

Ta có: \(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{75}< \dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{25}{50}=\dfrac{1}{2}\)

\(\dfrac{1}{76}+\dfrac{1}{77}+...+\dfrac{1}{100}< \dfrac{1}{75}+\dfrac{1}{75}+...+\dfrac{1}{75}=\dfrac{25}{75}=\dfrac{1}{3}\)

Do đó: \(A< \dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)(2)

Từ (1) và (2) ta suy ra ĐPCM

1/50+1/51+1/52+...+1/99<5/6<1/50.25+1/75.25=1/2+1/3=5/6(đpcm)

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