a: \(=\dfrac{4\cdot11\cdot3}{4\cdot9}\cdot\dfrac{1}{121}=\dfrac{11}{121}\cdot\dfrac{1}{3}=\dfrac{1}{33}\)
b: \(=\dfrac{7}{9}\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=\dfrac{7}{9}\)
c: \(=\dfrac{-6}{7}-\dfrac{1}{7}+\dfrac{2}{11}+\dfrac{9}{11}+\dfrac{12}{17}=\dfrac{12}{17}\)