Tìm x biết: x⁴ = 2x² – 12x + 8.

x⁴ = 2x² – 12x + 8.

⇔ x⁴ – 2x² + 12x – 8 = 0

⇔ (x⁴ + 2x³ – 2x²) + (–2x³ – 4x² + 4x) + (4x² + 8x – 8) = 0

⇔ x²(x² + 2x – 2) – 2x(x² + 2x – 2) + 4(x² + 2x – 2) = 0

⇔ (x² + 2x – 2)(x² – 2x + 4) = 0

Vậy $x=\sqrt{3}-1$ hoặc $x=-\sqrt{3}-1$.