\(\dfrac{6}{3.5}+\dfrac{6}{5.7}...+\dfrac{6}{33.35}\)
=\(3\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}...\dfrac{1}{55}+\dfrac{1}{55}-\dfrac{1}{57}\right)\)
=\(3.\left(\dfrac{1}{3}-\dfrac{1}{57}\right)\)
=\(3.\dfrac{6}{19}\)
=\(\dfrac{18}{19}\)