giải pt vô tỉ: $x^{2}-3x+1=-\frac{3}{\sqrt{3}}\sqrt{x^{4}+x^{2}+1}$
$\sqrt{2x^{2}-2x+1}+\sqrt{2x^{2}-(\sqrt{3}-1)x+1}+\sqrt{2x^{2}+(\sqrt{3}+1)x+1}=3$
$2x+\frac{x-1}{x}-\sqrt{1-\frac{1}{x}}-3\sqrt{x-\frac{1}{x}}=0$
$\sqrt{1-x}-2x\sqrt{1-x^{2}}-2x^{2}+1=0$