Giải thích các bước giải:
$(x^2+1)^2=5-x\sqrt{2x^2+4}$
$+)x>0$
$\to (x^2+1)^2=5-\sqrt{2x^4+4x^2}$
$\to (x^4+2x^2)-4=-\sqrt{2(x^4+2x^2)}$
$\to (x^4+2x^2)+\sqrt{2(x^4+2x^2)}-4=0$
$\to \sqrt{x^4+2x^2}=\sqrt{2}$ vì $ \sqrt{x^4+2x^2}\ge 0$
$\to x^4+2x^2=2$
$\to x^2=-1+\sqrt{3}, x^2\ge 0$
$\to x=\sqrt{-1+\sqrt{3}}, x>0$
$+)x<0$
$\to (x^2+1)^2=5+\sqrt{2x^4+4x^2}$
$\to (x^4+2x^2)-\sqrt{2(x^4+2x^2)}-4=0$
$\to (\sqrt{x^4+2x^2}-2\sqrt{2})(\sqrt{x^4+2x^2}+\sqrt{2})=0$
$\to \sqrt{x^4+2x^2}=2\sqrt{2}$
$\to x^4+2x^2=8$
$\to x^2=2$
$\to x=-\sqrt{2}, x<0$