\(\dfrac{1}{3\cdot10}+\dfrac{1}{10\cdot17}+...+\dfrac{1}{\left(7x-4\right)\left(7x+3\right)}=\dfrac{5}{114}\)
=>\(\dfrac{7}{3\cdot10}+\dfrac{7}{10\cdot17}+...+\dfrac{7}{\left(7x-4\right)\left(7x+3\right)}=\dfrac{35}{114}\)
=>\(\dfrac{1}{3}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{17}+...+\dfrac{1}{7x-4}-\dfrac{1}{7x+3}=\dfrac{35}{114}\)
=>\(\dfrac{1}{3}-\dfrac{1}{7x+3}=\dfrac{35}{114}\)
=>\(\dfrac{1}{7x+3}=\dfrac{1}{3}-\dfrac{35}{114}=\dfrac{1}{38}\)
=>7x+3=38
=>7x=35
=>\(x=\dfrac{35}{7}=5\)