a) x⁵ - 2x³ + x = 0 x·(x⁴ - 2x² + 1) = 0 x·[(x²)² - 2 · x² · 1 + 1²] = 0 x·(x² - 1)² = 0 x·[(x - 1)(x + 1)]² = 0 x·(x - 1)²(x + 1)² = 0 x = 0 ∨ (x - 1)² = 0 ∨ (x + 1)² = 0 x = 0 (x - 1)² = 0 x - 1 = 0 x = 1 (x + 1)² = 0 x + 1 = 0 x = - 1 Odp. x = - 1 lub x = 0 lub x = 1 b) x³ + 3x² + 2x = 0 x·(x² + 3x + 2) = 0 x = 0 ∨ x² + 3x + 2 = 0 x = 0 x² + 3x + 2 = 0 Δ = 3² - 4 · 1 · 2 = 9 - 8 = 1; √Δ = 1 x₁ = ⁻³⁻¹/₂·₁ = ⁻⁴/₂ = - 2 x₂ = ⁻³⁺¹/₂·₁ = ⁻²/₂ = - 1 Odp. x = - 2 lub x = - 1 lub x = 0 c) x⁴ = 4x³ + 5x² x⁴ - 4x³ - 5x² = 0 x²·(x² - 4x - 5) = 0 x² = 0 ∨ x² - 4x - 5 = 0 x² = 0 x = 0 x² - 4x - 5 = 0 Δ = (-4)² - 4 · 1 · (- 5) = 16 + 20 = 36; √Δ = 6 x₁ = ⁴ ⁻ ⁶ / ₂·₁ = ⁻ ²/₂ = - 1 x₂ = ⁴ ⁺ ⁶ / ₂·₁ =¹⁰/₂ = 5 Odp. Odp. x = - 1 lub x = 0 lub x = 5 d) 6x³ + 9x² = 3x⁴ -3x⁴ + 6x³ + 9x² = 0 - 3x²·(x² - 2x - 3) = 0 - 3x² = 0 ∨ x² - 2x - 3 = 0 - 3x² = 0 /:(-3) x² = 0 x = 0 x² - 2x - 3 = 0 Δ = (- 2)² - 4 · 1 · (- 3) = 4 + 12 = 16; √Δ = 4 x₁ = ² ⁻ ⁴/₂·₁ = ⁻ ²/₂ = - 1 x₂ = ²⁺ ⁴/₂·₁ = ⁶/₂ = 3 Odp. x = - 1 lub x = 0 lub x = 3 e) 2x⁵ = 2x⁴ + 12x³ 2x⁵ - 2x⁴ - 12x³ = 0 2x³·(x² - x - 6) = 0 2x³ = 0 ∨ x² - x - 6 = 0 2x³ = 0 /:2 x³ = 0 x = 0 x² - x - 6 = 0 Δ = (- 1)² - 4 · 1 · (- 6) = 1 + 24 = 25; √Δ = 5 x₁ = ¹ ⁻ ⁵/₂·₁ = ⁻ ⁴/₂ = - 2 x₂ = ¹ ⁺ ⁵/₂·₁ = ⁶/₂ = 3 Odp. x = - 2 lub x = 0 lub x = 3 f) 10x⁴ + x³ = 2x² 10x⁴ + x³ - 2x² = 0 x²·(10x² + x - 2) = 0 x² = 0 ∨ 10x² + x - 2 = 0 x² = 0 x = 0 10x² + x - 2 = 0 Δ = 1² - 4 · 10 · (- 2) = 1 + 80 = 81; √Δ = 9 x₁ = ⁻¹ ⁻ ⁹/₂·₁₀ = ⁻ ¹⁰/₂₀ = - ¹/₂ x₂ = ⁻¹ ⁺ ⁹/₂·₁₀ = ⁸/₂₀ = ²/₅ Odp. x = - ¹/₂ lub x = 0 lub x = ²/₅ g) 9x⁶ + 6x⁵ + x⁴ = 0 x⁴·(9x² + 6x + 1) = 0 x⁴·[(3x)² + 2·3x·1 + 1²] = 0 x⁴·(3x + 1)² = 0 x⁴ = 0 ∨ (3x + 1)² = 0 x⁴ = 0 x = 0 (3x + 1)² = 0 3x + 1 = 0 3x = - 1 /:3 x = - ¹/₃ Odp. x = - ¹/₃ lub x = 0 i) x³ + 4x = - 5x² x³ + 5x² + 4x = 0 x·(x² + 5x + 4) = 0 x = 0 ∨ x² + 5x + 4 = 0 x = 0 x² + 5x + 4 = 0 Δ = 5² - 4 · 1 · 4 = 25 - 16 = 9; √Δ = 3 x₁ = ⁻⁵ ⁻³/₂·₁ = ⁻ ⁸/₂ = - 4 x₂ = ⁻⁵ ⁺ ³/₂·₁ = ⁻²/₂ = - 1 Odp. x = - 4 lub x = - 1 lub x = 0 j) -½x⁴ + x³ = ½x² -½x⁴ + x³ - ½x² = 0 /·(-2) x⁴ - 2x³ + x² = 0 x²·(x² - 2x + 1) = 0 x²·(x² - 2·x·1 + 1²) = 0 x²·(x - 1)² = 0 x² = 0 ∨ (x - 1)² = 0 x² = 0 x = 0 (x - 1)² = 0 x - 1 = 0 x = 1 Odp. x = 0 lub x = 1 l) 16x⁷ + 8x⁵ + x³ = 0 x³·(16x⁴ + 8x² + 1) = 0 x³·[(4x²)² + 2·4x²·1 + 1²] = 0 x³·(4x² + 1)² = 0 x³ = 0 ∨ (4x² + 1)² = 0 x³ = 0 x = 0 (4x² + 1)² = 0 4x² + 1 = 0 Δ = 0² - 4 · 4 · 1 = 0 - 16 = - 16 < 0 równanie nie ma rozwiązań Odp. x = 0 * to co jest zapisane w rozwiązaniach kursywą nie trzeba przepisywać - pokazałam jak stosuje się wzór skróconego mnożenia
