Question

1)

Cho \(\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+...+\dfrac{1}{200}\)

Chứng minh: \(A>\dfrac{9}{10}\)

2)

Cho \(B=\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\)

Chứng minh: \(B>\dfrac{7}{12}\)

HELP ME!!!!!!!!

Answer 1

\(B=\left(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}\right)+\left(\dfrac{1}{151}+...\dfrac{1}{200}\right)>\dfrac{1}{150}+..\dfrac{1}{150}+\dfrac{1}{200}+..+200=\dfrac{50}{150}+\dfrac{50}{200}=\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{4}{12}+\dfrac{3}{12}=\dfrac{7}{12}\)Vậy ... (ta có điều phải chứng minh )

Answer 2

Ta có :\(\dfrac{1}{20}>\dfrac{1}{200}\)

...

\(\dfrac{1}{199}>\dfrac{1}{200}\)

Do đó : \(\dfrac{1}{20}+\dfrac{1}{21}+...+\dfrac{1}{200}>\dfrac{1}{200}+\dfrac{1}{200}+..+\dfrac{1}{200}=\dfrac{181}{200}>\dfrac{180}{200}=\dfrac{9}{10}\)Vậy ...