Bài 6:
\(B=\dfrac{1}{199}+1+\dfrac{2}{198}+1+\dfrac{3}{197}+1+...+\dfrac{198}{2}+1+1\)
\(=\dfrac{200}{199}+\dfrac{200}{198}+...+\dfrac{200}{2}+\dfrac{200}{200}\)
\(=200\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{200}\right)=200\cdot A\)
=>A/B=1/200