\(B=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\frac{99}{1}+\frac{98}{2}+...+\frac{2}{98}+\frac{1}{99}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\frac{98}{2}+1+\frac{97}{3}+1+...+\frac{2}{98}+1+\frac{1}{99}+1}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\frac{100}{2}+\frac{100}{3}+...+\frac{100}{98}+\frac{100}{99}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{100\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)}=\frac{1}{100}\)
A= 99/1+98/2+...+2/98+1/99
<=>A= (99/1-98)+(98/2+1)+....+(2/98+1)+(1/99+1)
<=>A= 100/100+100/2+...+100/98+100/99
A= 100( 1/100+1/2+...+1/98+1/99)
Vậy B=1/100
-----------------------Good luck-------------------