a) `(x-2)(x-1)=x(2x+1)+2`
⇔`x^2-x-2x+2=2x^2+x+2`
⇔`x^2-3x+2-2x^2-x-2=0`
⇔`-x^2-4x=0`
⇔`-x(x+4)=0`
⇔\(\left[ \begin{array}{l}-x=0\\x+4=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=0\\x=-4\end{array} \right.\)
Vậy `S={0,-4}`
b) `(x+2)(x+2)-(x-2)(x-2)=8x`
⇔`(x+2)^2-(x-2)^2=8x`
⇔`x^2+4x+4-x^2+4x-4-8x=0`
⇔`0x=0`
Vậy phương trình có vô số nghiệm
c) `(2x-1)(x^2-x+1)=2x^3-3x^2+2`
⇔`2x^3-2x^2+2x-x^2+x-1=2x^3-3x^2+2`
⇔`2x^3-2x^3-3x^2+3x^2+3x-1-2=0`
⇔`3x-3=0`
⇔`3x=3`
⇔`x=1`
Vậy `S={1}`
