Odpowiedź :
1) x³ + 6x² + 6x > -5
x³ + 6x² + 6x > -5
x³ + 6x² + 6x + 5 > 0
--- | 1 | 6 | 6| 5
-1 | 1 | 5 | 1 | 4
-5 | 1 | 1 | 1 | 0
(x + 5)(x² + x + 1) > 0
Δ = 1 - 4 < 0
x² + x + 1 > 0 (zawsze)
x > -5
2) - x³ + 2x² + 4x > 3
- x³ + 2x² + 4x > 3
- x³ + 2x² + 4x - 3 > 0
x³ - 2x² - 4x + 3 < 0
- | 1 | -2 | -4 | 3
1 | 1 | -1 | -5 | -2
-1| 1 | -3 | -1 | 4
3 | 1 | 1 | -1 | 0
(x - 3)(x² + x - 1) < 0
Δ = 1 + 4 = 5
√Δ = √5
x₁ = (-1 + √5)/2 = (√5 - 1)/2
x₁ = (-1 - √5)/2
(x - 3)(x - (√5 - 1)/2)(x - (-1 - √5)/2) < 0
x∈ ((-1 - √5)/2, (√5 - 1)/2) u (3, ω)
3) x³ - 6x² + 12x ≤ 8
x³ - 6x² + 12x ≤ 8
x³ - 6x² + 12x - 8 ≤ 0
-- | 1 | -6 | 12 | -8
1 | 1 | -5 | 7 | -1
2 | 1 | -4 | 4 | 0
(x - 2)(x² - 4x + 4) ≤ 0
(x - 2)(x - 2)² ≤ 0
x ≤ 2
x³ + 6x² + 6x > -5
x³ + 6x² + 6x + 5 > 0
--- | 1 | 6 | 6| 5
-1 | 1 | 5 | 1 | 4
-5 | 1 | 1 | 1 | 0
(x + 5)(x² + x + 1) > 0
Δ = 1 - 4 < 0
x² + x + 1 > 0 (zawsze)
x > -5
2) - x³ + 2x² + 4x > 3
- x³ + 2x² + 4x > 3
- x³ + 2x² + 4x - 3 > 0
x³ - 2x² - 4x + 3 < 0
- | 1 | -2 | -4 | 3
1 | 1 | -1 | -5 | -2
-1| 1 | -3 | -1 | 4
3 | 1 | 1 | -1 | 0
(x - 3)(x² + x - 1) < 0
Δ = 1 + 4 = 5
√Δ = √5
x₁ = (-1 + √5)/2 = (√5 - 1)/2
x₁ = (-1 - √5)/2
(x - 3)(x - (√5 - 1)/2)(x - (-1 - √5)/2) < 0
x∈ ((-1 - √5)/2, (√5 - 1)/2) u (3, ω)
3) x³ - 6x² + 12x ≤ 8
x³ - 6x² + 12x ≤ 8
x³ - 6x² + 12x - 8 ≤ 0
-- | 1 | -6 | 12 | -8
1 | 1 | -5 | 7 | -1
2 | 1 | -4 | 4 | 0
(x - 2)(x² - 4x + 4) ≤ 0
(x - 2)(x - 2)² ≤ 0
x ≤ 2
a) x² * x + 6x² + 6x > -5
7x² + 7x > -5
7 (x * x) + 7x > -5
7x + 7x + 7x > -5
21 x > -5
x > -5/21
7x² + 7x > -5
7 (x * x) + 7x > -5
7x + 7x + 7x > -5
21 x > -5
x > -5/21