<=> \(\left(x^2+1\right)+3x-x\sqrt{x^2+1}-3\sqrt{x^2+1}=0\)

<=> \(\left(\left(x^2+1\right)-3\sqrt{x^2+1}\right)+\left(3x-x\sqrt{x^2+1}\right)=0\)

<=> \(\sqrt{x^2+1}\left(\sqrt{x^2+1}-3\right)+x\left(3-\sqrt{x^2+1}\right)=0\)

<=> \(\left(\sqrt{x^2+1}-3\right)\left(\sqrt{x^2+1}-x\right)=0\)

<=> \(\sqrt{x^2+1}-3=0\) hoặc \(\sqrt{x^2+1}-x=0\)

+) \(\sqrt{x^2+1}-3=0\) => x² + 1 = 9 <=> x² = 8 <=> \(x=2\sqrt{2}\) hoặc \(x=-2\sqrt{2}\)

+) \(\sqrt{x^2+1}-x=0\)<=> \(x^2+1=x^2\) và x > 0  <=> 1 = 0 (Vô lí)

Vậy....