\(\dfrac{1}{11}\cdot\left(17-3\sqrt{x-1}\right)=\dfrac{1}{15}\cdot\left(23-4\sqrt{x-1}\right)\)

Condition \(x\ge1\)

\(\Leftrightarrow\dfrac{17}{11}-\dfrac{3\sqrt{x-1}}{11}=\dfrac{23}{15}-\dfrac{4\sqrt{x-1}}{15}\)

\(\Leftrightarrow\dfrac{17}{11}-\dfrac{23}{15}=-\dfrac{4\sqrt{x-1}}{15}+\dfrac{3\sqrt{x-1}}{11}\)

\(\Leftrightarrow\dfrac{2}{165}=\dfrac{-44\sqrt{x-1}+45\sqrt{x-1}}{165}\)

\(\Rightarrow2=\sqrt{x-1}\)

\(\Leftrightarrow4=x-1\Rightarrow x=5\) (satisfy with condition)

Hence, \(S=\left\{5\right\}\)