\(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)

\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{x\left(x+3\right)}\right)=\dfrac{101}{1540}\)

\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)

\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)

\(\Leftrightarrow\dfrac{1}{x+3}=\dfrac{1}{308}\)\(\Leftrightarrow x+3=308\Rightarrow x=305\)

\(\Rightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)

\(\Rightarrow\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{101}{1540}\div\dfrac{1}{3}\)

\(\Rightarrow\dfrac{x+3}{5x+15}-\dfrac{5}{5x+15}=\dfrac{303}{1540}\)

\(\Rightarrow\dfrac{x+3-5}{5x+15}=\dfrac{303}{1540}\)

\(\Rightarrow\dfrac{x-2}{5x+15}=\dfrac{303}{1540}\)

=> 1540x - 3080 = 1515x + 4545

=> 1540x - 1515x = 4545 + 3080

=> 25x = 7625

=> x = 305