Bài 2:
a: \(=6+\dfrac{4}{9}+3+\dfrac{7}{11}-4-\dfrac{4}{9}=5+\dfrac{7}{11}=\dfrac{62}{11}\)
b: \(=\dfrac{-4}{3}-\dfrac{17}{6}\cdot\dfrac{6}{11}+3:\dfrac{1}{20}\)
=-4/3-17/11+60
=1885/33
c: \(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}=\dfrac{1}{3}-\dfrac{1}{99}=\dfrac{32}{99}\)