Ta có: \(A=\frac12\cdot\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\cdot\ldots\cdot\left(1+\frac{1}{2021\cdot2023}\right)\)
\(=\frac12\left(1+\frac{1}{2^2-1}\right)\left(1+\frac{1}{3^2-1}\right)\cdot\ldots\cdot\left(1+\frac{1}{2022^2-1}\right)\)
\(=\frac12\cdot\frac{2^2-1+1}{2^2-1}\cdot\frac{3^2-1+1}{3^2-1}\cdot\ldots\cdot\frac{2022^2-1+1}{2022^2-1}\)
\(=\frac12\cdot\frac{2^2}{2^2-1}\cdot\frac{3^2}{3^2-1}\cdot\ldots\cdot\frac{2022^2}{2022^2-1}\)
\(=\frac12\cdot\frac{2^2}{1\cdot3}\cdot\frac{3^2}{2\cdot4}\cdot\ldots\cdot\frac{2022^2}{2021\cdot2023}=\frac12\cdot\frac{2\cdot3\cdot\ldots\cdot2022}{1\cdot2\cdot\ldots\cdot2021}\cdot\frac{2\cdot3\cdot\ldots\cdot2022}{3\cdot4\cdot\ldots\cdot2023}\)
\(=\frac12\cdot2022\cdot\frac{2}{2023}=\frac{2022}{2023}\)
