Cho hàm số $f\left( x \right)=\ln \dfrac{2018x}{x+1}$. Tính tổng $S={f}'\left( 1 \right)+{f}'\left( 2 \right)+...+{f}'\left( 2018 \right)$.

Ta có $f'\left( x \right)=\dfrac{2018}{{{\left( x+1 \right)}^{2}}}.\dfrac{x+1}{2018x}=\dfrac{1}{x\left( x+1 \right)}=\dfrac{1}{x}-\dfrac{1}{x+1}$
Ta có
$S=f'\left( 1 \right)+f'\left( 2 \right)+f'\left( 3 \right)+...+f'\left( 2018 \right)$
$=\left( 1-\dfrac{1}{2} \right)+\left( \dfrac{1}{2}-\dfrac{1}{3} \right)+\left( \dfrac{1}{3}-\dfrac{1}{4} \right)+...+\left( \dfrac{1}{2018}-\dfrac{1}{2019} \right)$
$=1-\dfrac{1}{2019}=\dfrac{2018}{2019}.$