\(\left(3x-2\right)\left(x+1\right)=3x\left(x-3\right)\)
\(\Leftrightarrow3x^2+3x-2x-2=3x^2-9x\)
\(\Leftrightarrow3x^2-3x^2+3x-2x+9x=2\)
\(\Leftrightarrow10x=2\)
\(\Leftrightarrow x=\dfrac{1}{5}\)
Vậy \(S=\left\{\dfrac{1}{5}\right\}\)
(3x-2)(x+1)=3x(x-3)
⇔3x2+3x-2x-2=3x2-9x
⇔3x2-3x2+x-9x=2
⇔-8x=-2
⇔x=4
vậy pt có tập no S={4}