ĐK: $x\geq 0$
$PT\Leftrightarrow (x-1-\sqrt{x})+(\sqrt{(x^2-x+1)(x^2-3x+3)}-\sqrt{x^2+x+1})=0$
$(x^2-3x+1)(\frac{1}{x-1+\sqrt{x}}+\frac{x^2-x+2}{\sqrt{(x^2-x+1)(x^2-3x+3)}+\sqrt{x^2+x+1}})=0$
Mà $(....)>0$ nên $x^2-3x+1=0$
$x=\frac{3+\sqrt{5}}{2};x=\frac{3-\sqrt{5}}{2}$(tm)