Condition : \(x\ne3;-5\)

\(\dfrac{x+5}{3}-\dfrac{x-3}{5}=\dfrac{5}{x-3}-\dfrac{3}{x+5}\)

\(\Leftrightarrow\dfrac{5\left(x+5\right)-3\left(x-3\right)}{15}=\dfrac{5\left(x+5\right)-3\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}\)

\(\Leftrightarrow\left(2x+34\right)\left(\dfrac{1}{15}-\dfrac{1}{\left(x-3\right)\left(x+5\right)}\right)=0\)

\(\Leftrightarrow\left(x+17\right).\dfrac{\left(x-3\right)\left(x+5\right)-15}{15\left(x-3\right)\left(x+5\right)}=0\)

\(\Leftrightarrow\left(x+17\right)\left(x^2+2x-30\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+17=0\\x^2+2x-30=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-17\\x=-1\pm\sqrt{31}\end{matrix}\right.\)(satisfied)

So, \(S=\left\{-17;-1\pm\sqrt{31}\right\}\)

Condition : x≠3;−5

x+53−x−35=5x−3−3x+5

⇔5(x+5)−3(x−3)15=5(x+5)−3(x−3)(x−3)(x+5)

⇔(2x+34)(115−1(x−3)(x+5))=0

⇔(x+17).(x−3)(x+5)−1515(x−3)(x+5)=0

⇔(x+17)(x2+2x−30)=0

⇒[x+17=0x2+2x−30=0

⇒[x=−17x=−1±√31

(satisfied)

So, S={−17;−1±√31}