\(\left(3-2x\right)^2=\left(x-2\right)\left(2x-3\right)\)
\(\Leftrightarrow\left(3x-2\right)^2-\left(x-2\right)\left(2x-3\right)=0\)
\(\Leftrightarrow9x^2-12x+4-\left(2x^2-7x+6\right)=0\)
\(\Leftrightarrow9x^2-12x+4-2x^2+7x-6=0\)
\(\Leftrightarrow7x^2-5x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{7}\end{matrix}\right.\)
Vậy \(S=\left\{1;-\dfrac{2}{7}\right\}\)
`(3-2x)^2=(x-2)(2x-3)`
`<=>(2x-3)^2 -(x-2)(2x-3)=0`
`<=> (2x-3)(2x-3-x+2)=0`
`<=> (2x-3)(x-1)=0`
\(< =>\left[{}\begin{matrix}2x-3=0\\x-1=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=1\end{matrix}\right.\)
