Question

 

 

Answer 1

\(\left(x+2\right)\left(\dfrac{120}{x}+1\right)=160\left(x\ne0\right)\\ < =>120+x+\dfrac{240}{x}+2=160\\ < =>x+\dfrac{240}{x}-38=0\\ < =>\dfrac{x^2+240-38x}{x}=0\\ < =>x^2-38x+240=0\\ < =>\left(x^2-38x+19^2\right)-121=0\\ < =>\left(x-19\right)^2=121=\left(\pm11\right)^2\\ =>\left[{}\begin{matrix}x-19=11\\x-19=-11\end{matrix}\right.< =>\left[{}\begin{matrix}x=30\\x=8\end{matrix}\right.\left(TMDK\right)=>S=\left\{30;8\right\}\)