\(\dfrac{3}{x}=\dfrac{5}{x-3}\left(x\ne0;x\ne3\right)\) đề như thế này phải ko ạ?
suy ra: \(3\left(x-3\right)=5x\\ < =>3x-9=5x\\ < =>3x-5x=9\\ < =>-2x=9\\ < =>x=-\dfrac{9}{2}\left(tm\right)\)
\(\dfrac{3}{x}=\dfrac{5}{x-3}\)
ĐKTC: \(x\ne0\)
\(x\ne3\)
\(\dfrac{3\left(x-3\right)}{x\left(x-3\right)}=\dfrac{5x}{x\left(x-3\right)}\)
Suy ra: \(3\left(x-3\right)=5x\)
<=> \(3x-9-5x=0\)
<=> \(-2x-9=0\)
<=> \(-2x=9\)
<=> \(x=-\dfrac{9}{2}=-4,5\)
Vậy \(S=\left\{-4,5\right\}\)