Ta có: \(A=\dfrac{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{9}}{\dfrac{4}{3}+\dfrac{4}{5}-\dfrac{4}{9}}\)
\(=\dfrac{2\left(\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{9}\right)}{4\left(\dfrac{1}{3}+\dfrac{1}{5}-\dfrac{1}{9}\right)}\)
\(=\dfrac{2}{4}=\dfrac{1}{2}.\)
Vậy \(A=\dfrac{1}{2}.\)