\((x^2+x+1)^2-2x^2-2x=5\\↔(x^2+x+1)^2-2x^2-2x-2-3=0\\↔(x^2+x+1)^2-2(x^2+x+1)-3=0\\ Dat\,\, x^2+x+1=t\\→t^2-2t-3=0\\↔t^2-3t+t-3=0\\↔t(t-3)+(t-3)=0\\↔(t+1)(t-3)=0\\↔t+1=0\quad or\quad t-3=0\\↔t=-1\quad or\quad t=3\\↔x^2+x+1=-1\quad or\quad x^2+x+1=3\\↔x^2+x+2=0\quad or\quad x^2+x-2=0\\↔x^2+2x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{7}{4}=0\quad or\quad x^2+2x-x-2=0\\↔\bigg(x+\dfrac{1}{2}\bigg)^2+\dfrac{7}{4}=0(vo\,\, ly)\quad or\quad x(x+2)-(x+2)=0\\↔(x-1)(x+2)=0\\↔x-1=0\quad or\quad x+2=0\\↔x=1\quad or\quad x=-2\)
Vậy pt có tập nghiệm \(S=\{-2;1\}\)
