Question

Giúp bài 1,2,3,4,5 nha

Answer 1

Bài 5:

a: \(A=3\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=3\cdot\dfrac{99}{100}=\dfrac{297}{100}\)

b: \(B=\dfrac{3}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{98}-\dfrac{1}{100}\right)\)

\(=\dfrac{3}{2}\cdot\dfrac{49}{100}=\dfrac{147}{200}\)