Question

Tính \(\frac{A}{B}\)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

Ta có: \(B=\frac{1}{199}+\frac{2}{198}+...+\frac{199}{1}\)

\(=\frac{200-199}{199}+\frac{200-198}{198}+...+\frac{200-1}{1}\)

\(=\frac{200}{199}-\frac{199}{199}+\frac{200}{198}-\frac{198}{198}+...+\frac{200}{1}-\frac{1}{1}\)

\(=\left(\frac{200}{199}+\frac{200}{198}+...+\frac{200}{1}\right)-\left(\frac{199}{199}+\frac{198}{198}+...+\frac{1}{1}\right)\)

\(=200+200\left(\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)-199\)

\(=200\left(\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)+\frac{200}{200}\)

\(=200\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)\)

\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}}{200\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)}=\frac{1}{200}\)

Answer 2

Ta có :

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

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

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

\(B=200\left(\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)\)

\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}}{200\left(\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)}=\frac{1}{200}\)

Vậy \(\frac{A}{B}=\frac{1}{200}\)

Answer 3

Ta có : 

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

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

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

\(B=200\left(\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)\)

\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}}{200\left(\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)}=\frac{1}{200}\)

Vậy \(\frac{A}{B}=\frac{1}{200}\)

Answer 4

Ta có : 

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

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

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

\(B=200\left(\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)\)

\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}}{200\left(\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)}=\frac{1}{200}\)

Vậy \(\frac{A}{B}=\frac{1}{200}\)

Answer 5

Ta có : 

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

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

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

\(B=200\left(\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)\)

\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}}{200\left(\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)}=\frac{1}{200}\)

Vậy \(\frac{A}{B}=\frac{1}{200}\)