\(\left(3x-1\right)\left(x+3\right)=\left(2-x\right)\left(5-3x\right)\)
\(\Leftrightarrow3x^2+8x-3=10-11x+3x^2\)
\(\Leftrightarrow19x=13\)
\(\Leftrightarrow x=\dfrac{13}{19}\)
\(\Rightarrow S=\left\{\dfrac{13}{19}\right\}\)
\(\left(3x-1\right)\left(x+3\right)=\left(2-x\right)\left(5-3x\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(x+3\right)-\left(2-x\right)\left(5-3x\right)=0\)
\(\Leftrightarrow3x^2+8x-3-10+11x-3x^2=0\)
\(\Leftrightarrow19x-13=0\Leftrightarrow19x=13\Leftrightarrow x=\dfrac{13}{19}\)
Vậy, \(S=\left\{\dfrac{13}{19}\right\}\)