Question

1. |x₃+2x|-3x >0

2. |x⁴-1|-3(x²-1)≥0

Answer 1

1. |x³ + 2x| - 3x > 0
|x(x² + 2)| - 3x > 0
x² + 2 > 0 zawsze:
(x² + 2)|x| - 3x > 0

dla x ≥ 0
x(x² + 2) - 3x > 0
x(x² + 2 - 3) > 0
x(x² - 1) > 0
x(x - 1)(x + 1) > 0
x ∈ (-∞, - 1) u (0, 1) i x ≥ 0 => x ∈ (0, 1)

dla x < 0:
- x(x² + 2) - 3x > 0
x(x² + 2 + 3) < 0
x(x² + 5) < 0
Δ < 0, x² + 5 > 0
x < 0 i x < 0 => x < 0

x ∈ (0, 1) lub x < 0
x ∈ (-∞, 0) u (0, 1)


2. |x⁴ - 1| - 3(x² - 1) ≥ 0
|(x² - 1)(x² + 1)| - 3(x² - 1) ≥ 0
x² + 1 > 0
(x² + 1)|(x - 1)(x + 1)| - 3(x² - 1) ≥ 0

dla x ∈ (- 1, 1)
- (x² + 1)(x - 1)(x + 1) - 3(x² - 1) ≥ 0
(x² - 1)(x² + 1 + 3) ≤ 0
(x - 1)(x + 1) ≤ 0
x ∈ (- 1, 1) i x ∈ <- 1, 1> => x ∈ (- 1, 1)

dla x ∈ (-∞, - 1> u <1, ∞)
(x² + 1)(x - 1)(x + 1) - 3(x² - 1) ≥ 0
(x² - 1)(x² + 1 - 3) ≥ 0
(x - 1)(x + 1)(x - √2)(x + √2) ≥ 0
x ∈ (- ∞, - √2> u <- 1, 1> u <√2, ∞) i x ∈ (-∞, - 1> u <1, ∞) => x ∈ (- ∞, - √2> u {- 1, 1} u <√2, ∞)

x ∈ (- 1, 1) i x ∈ (- ∞, - √2> u {- 1, 1} u <√2, ∞) => x ∈ (- ∞, - √2> u <- 1, 1> u <√2, ∞)

jak masz pytania to pisz na pw