2,
a, $(x+2)^2=x+2$
$⇔(x+2)^2-(x+2)=0$
$⇔(x+2)(x+2-1)=0$
$⇔(x+2)(x+1)=0$
$⇔x+2=0$ hoặc $x+1=0$
$x=-2$ $x=-1$
Vậy `S={-2,-1}`
b, $x^3+4x=0$
$⇔x(x^2+4)=0$
$⇔x=0$ (do `x^2+4` luôn $≥4>0)$
Vậy `S={0}`
c, $(2x+3)(x-1)+(2x-3)(1-x)=0$
$⇔(2x+3)(x-1)-(2x-3)(x-1)=0$
$⇔(x-1)(2x+3-2x+3)=0$
$⇒6(x-1)=0$
$⇔x-1=0$
$⇔x=1$
Vậy `S={1}`