[tex]D_f:x\not =0 \wedge x\not=-3 \\
D_f:x\in\mathbb{R}\setminus\{-3,0\}\\
\frac{17}{x}-\frac{11}{x+3}=\frac{5x+8}{x+3}|\cdot x(x+3)\\
17(x+3)-11x=x(5x+8)\\
17x+51-11x=5x^2+8x\\
5x^2+2x-51=0\\
5x^2-15x+17x-51=0\\
5x(x-3)+17(x-3)=0\\
(5x+17)(x-3)=0\\
x=-\frac{17}{5} \vee x=3
[/tex]
