Bài 1:
a) `2x^3-3x^2+4x-3=0`
`⇔2x^3-2x^2-x^2+x+3x-3=0`
`⇔2x^2(x-1)-x(x-1)+3(x-1)=0`
`⇔(x-1)(2x^2-x+3)=0`
`⇔x-1=0` (vì `2x^2-x+3>0`)
`⇔x=1`
Vậy `S={1}`
b) `x^3-3x+2=0`
`⇔x^3-x-2x+2=0`
`⇔x(x^2-1)-2(x-1)=0`
`⇔x(x-1)(x+1)-2(x-1)=0`
`⇔(x-1)(x^2+x-2)=0`
`⇔(x-1)(x^2-x+2x-2)=0`
`⇔(x-1)[x(x-1)+2(x-1)]=0`
`⇔(x-1)^2(x+2)=0`
`⇔`\(\left[ \begin{array}{l}(x-1)^2=0\\(x+2=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=1\\x=-2\end{array} \right.\)
Vậy `S={1;-2}`
