3Хчетвертой степени-4Хв квадрате + 1=0

[latex]3x^4 - 4x^2 +1 = 0 \ | \ t= x^2 \\ 3t^2 - 4t +1 =0 \\ t_{1,2} = 1, \frac13 \Rightarrow \\ 1. \ \ x^2 = 1 \Rightarrow x_{1,2} = \pm 1. \\ 2. \ \ x^2 = \frac13 \Rightarrow x_{3,4} = \pm \frac1{\sqrt{3}}[/latex] Ответ: [latex]1, -1, \frac1{\sqrt3}, -\frac1{\sqrt3}.[/latex]