a) x2-x=-2x+2
<=>x(x-1)=-2(x-1)
<=>x(x-1)+2(x-1)=0
<=> (x-1)(x+2)=0
=> x-1=0 hoặc x+2=0
=> x =1 hoặc x =-2
Vậy S={-2;1).
b)4x2+4x+1=x2
<=>(2x+1)2-x2=0
<=>(2x+1+x)(2x+1-x)=0
<=>(3x+1)(x+1)=0
=> 3x+1=0 hoặc x+1=0
=> 3x =-1 x=-1
=> x=\(-\dfrac{1}{3}\) x=-1
Vây S={\(-\dfrac{1}{3}\);-1}