Question

tính tỉ số \(\frac{A}{B}\)biết:

A=\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}\)

B=\(\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}\)

Answer 1

B = \(\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+\frac{199}{1}\)

\(=\frac{1}{199}+\frac{2}{198}+\frac{3}{197}+...+\frac{198}{2}+199\)

\(=\left(\frac{1}{199}+1\right)+\left(\frac{2}{198}+1\right)+...+\left(\frac{198}{2}+1\right)+1\)

(từ 1 đến 198 có 198 số hạng nên còn 1 số 1)

\(=\frac{200}{199}+\frac{200}{198}+...\frac{200}{2}+\frac{200}{200}\)

\(=200\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{\text{4}}+...+\frac{1}{200}\right)=200A\)

=> B = 200A => \(\frac{A}{B}=\frac{1}{200}\)

Vậy \(\frac{A}{B}=\frac{1}{200}\)