\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{17\times18}+\dfrac{1}{18\times19}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{17}-\dfrac{1}{18}+\dfrac{1}{18}-\dfrac{1}{19}\\ =1-\dfrac{1}{19}\\ =\dfrac{18}{19}\)