Rozwiąż nieróność
g) 2x⁴+x²-3≥(1-2x²)(3-x²)
2x⁴+x²-3≥(1-2x²)(3-x²)
2x⁴+x²-3 - (1-2x²)(3-x²) ≥0
2x⁴+x²-3 - (2x² - 1)(x² -3) =0
zamiana t=x²
2t² + t - 3 - (2t - 1)(t -3) = 0
2t² + t - 3 - 2t² +7t -3 = 0
8t - 6 = 0
8t = 6
t = 0,75
zamiana t=x² ⇒ x² = 0,75 ⇒ x = ± 1/2*√3
Odp: x∈ (-∞; -1/2*√3] ∪ [ 1/2*√3; +∞)
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2x⁴+x²-3≥(1-2x²)(3-x²)
2x⁴+x²-3 - (1-2x²)(3-x²) ≥0
2x⁴+x²-3 - (2x² - 1)(x² -3) =0
zamiana t=x²
2t² + t - 3 - (2t - 1)(t -3) = 0
2t² + t - 3 - 2t² +7t -3 = 0
8t - 6 = 0
8t = 6
t = 0,75
zamiana t=x² ⇒ x² = 0,75 ⇒ x = ± 1/2*√3
Odp: x∈ (-∞; -1/2*√3] ∪ [ 1/2*√3; +∞)