2:
a: \(=\dfrac{2}{5}+\dfrac{4}{21}=\dfrac{42+20}{105}=\dfrac{62}{105}\)
b: \(=\dfrac{2}{6}\cdot\dfrac{5}{7}+\dfrac{2}{6}=\dfrac{1}{3}\left(\dfrac{5}{7}+1\right)=\dfrac{1}{3}\cdot\dfrac{12}{7}=\dfrac{4}{7}\)
c: \(=\left(\dfrac{4}{6}-\dfrac{1}{6}\right)\cdot\dfrac{6}{7}=\dfrac{3}{6}\cdot\dfrac{6}{7}=\dfrac{3}{7}\)