Решите уравнение 4x^4-5x^2+1=0.

\[4x^{4} - 5x^{2} + 1 = 0\]

\[Пусть\ x^{2} = t \geq 0:\]

\[4t^{2} - 5t + 1 = 0\]

\[D = 25 - 16 = 9\]

\[t_{1} = \frac{5 + 3}{8} = 1;\ \ t_{2} = \frac{5 - 3}{8} = \frac{2}{8} = \frac{1}{4}.\]

\[1)\ x^{2} = 1\]

\[x = \pm 1.\]

\[2)\ x^{2} = \frac{1}{4}\]

\[x = \pm \frac{1}{2}.\]

\[Ответ:x = \pm 1;\ \ x = \pm \frac{1}{2}.\]