Question

[Trần Thị Hương Lan]

Chứng minh rằng:

\(A=\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{199}+\dfrac{1}{200}>\dfrac{7}{12}\)

Giúp mk nhé!

Answer 1

[阮草~๖ۣۜDαɾƙ]

Ta có:

\(\dfrac{1}{101}>\dfrac{1}{150}\)

\(\dfrac{1}{102}>\dfrac{1}{150}\)

....

\(\dfrac{1}{150}=\dfrac{1}{150}\)

=>\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{150}>\dfrac{1}{150}+\dfrac{1}{150}+...+\dfrac{1}{150}\)(50 số)=\(\dfrac{1}{3}\)

Ta có:

\(\dfrac{1}{152}>\dfrac{1}{200}\)

\(\dfrac{1}{153}>\dfrac{1}{200}\)

....

\(\dfrac{1}{200}=\dfrac{1}{200}\)

=>\(\dfrac{1}{151}+\dfrac{1}{153}+...+\dfrac{1}{120}>\dfrac{1}{120}+\dfrac{1}{120}+...+\dfrac{1}{120}\)(50 số)=\(\dfrac{1}{4}\)

=>\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{200}>\dfrac{1}{3}+\dfrac{1}{4}\)

=> \(A>\dfrac{7}{12}\)

Answer 2

[Trần Thị Hương Lan]

Cảm ơn bạn.