\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{98x99}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{99}\)
\(=\frac{1}{2}-\frac{1}{99}\)
\(=\frac{99}{198}-\frac{2}{198}\)
\(=\frac{97}{198}\)
\(\frac{A}{198}=\frac{97}{198}=>A=198x97:198=97\)