\(A=\frac{1}{2.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{50}\)
\(\Rightarrow A=\frac{12}{25}\)
Vậy \(A=\frac{12}{25}\)