\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...-\frac{1}{101}\)
\(A=\left(\frac{1}{3}-\frac{1}{101}\right):2\)= 49/303
\(A=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+...+\frac{1}{99\times101}\)
\(=\frac{1}{2}\times\left(\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}+...+\frac{2}{99\times101}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{101}\right)\)
\(=\frac{1}{2}\times\frac{98}{303}=\frac{49}{303}\)
Vậy A = 49/303.
