[tex]4\sqrt{x^{2}+x+1}=1+5x+4x^{2}-2x^{3}-x^{4}[/tex]
Đặt t=căn x^2+x+1 ;t>=căn3/2
t^2=x^2+x+1<=>5t^2-5=5x^2+5x
(t^2-1)^2=(x^2+x)^2
<=>t^4-2t^2+2=x^4+2x^3+x^2
<=>-t^4+2t^2-2=-x^4-2x^3-x^2
Pt<=>4t=7t^2-t^4-5
<=>(t^2-3)^2-(t-2)^2=0
<=>(t^2-t-1)(t^2+t-5)=0