\(\left(3x-2\right)\left(2x+1\right)=\left(2x-1\right)^2\)
=>\(6x^2+3x-4x-2=4x^2-4x+1\)
=>\(6x^2-x-2-4x^2+4x-1=0\)
=>\(2x^2+3x-3=0\)
\(\Delta=3^2-4\cdot2\cdot\left(-3\right)=9+24=33>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left[{}\begin{matrix}x=\dfrac{-3-\sqrt{33}}{2\cdot2}=\dfrac{-3-\sqrt{33}}{4}\\x=\dfrac{-3+\sqrt{33}}{4}\end{matrix}\right.\)
\(\left(3x-2\right)\left(2x+1\right)=\left(2x-1\right)^2\)
\(\Leftrightarrow6x^2+3x-4x-2=4x^2-4x+1\)
\(\Leftrightarrow6x^2-4x^2+3x-4x+4x-2-1=0\)
\(\Leftrightarrow2x^2+3x-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3+\sqrt{33}}{4}\\x=\dfrac{-3-\sqrt{33}}{4}\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{-3+\sqrt{33}}{4};\dfrac{-3-\sqrt{33}}{4}\right\}\).
