`i)(3x+2)(x^2-1)=(9x^2-4)(x+1)`
`⇔(3x+2)(x-1)(x+1)=(3x+2)(3x-2)(x+1)`
`⇔x-1=3x-2`
`⇔x-3x=-2+1`
`⇔-2x=-1`
`⇔x=1/2`
Vậy `S={1/2}`
`k)(x^2-4)-(x-2)(3-2x)=0`
`⇔(x-2)(x+2)-(x-2)(3-2x)=0`
`⇔(x-2)(x+2-3+2x)=0`
`⇔(x-2)(3x-1)=0`
`⇔` \(\left[ \begin{array}{l}x-2=0\\3x-1=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=2\\x=\dfrac{1}{3}\end{array} \right.\)
Vậy `S={2;1/3}`
