Question

Znajdź liczby spełniające równanie:
1) x³(x³-1)(1+x³)=0
2) (x³+2x)(x³+2)(x³+x)=0
3) (4x²-8x+6)(4x²-8x)(-8x+6)=0

Answer 1

1) x³(x³-1)(1+x³)=0
x³=0 x=0
x³-1=0 x³=1 x=1
x³+1=0 x³=-1 x=-1

2) (x³+2x)(x³+2)(x³+x)=0
x³+2x=0 x(x²+2)=0 x₁=0 ∨ x²+2=0 x²=-2
x³+2=0 x³=-2 x₂=-∛2
x³+x=0 x(x²+1)=0 x=0 x²=-1


3) (4x²-8x+6)(4x²-8x)(-8x+6)=0
a=4 b=-8 c=6 Δ=64-96 Δ=-32
4x²-8x=0 4x(x-2)=0 x₁=0 ∨ x₂=2
-8x+6=0 8x=6 x₃=¾

Answer 2

1) x³(x³-1)(1+x³)=0
x³(x-1)(x²+x+1)(x+1)(x²-x+1) = 0
=> x=0 ∨ x=1 ∨ x=-1

2) (x³+2x)(x³+2)(x³+x)=0
x(x²+2)(x+∛2)(x²-x+(∛2)²)x(x²+1) = 0
=> x=0 ∨ x=-∛2

3) (4x²-8x+6)(4x²-8x)(-8x+6)=0
(4x²-8x+6)4x(x-2)(-8x+6) = 0
=> x=0 ∨ x=2 ∨ x=6/8

Answer 3

1) x³(x³ - 1)(1 + x³) = 0
x³(x - 1)(x² + x + 1)(x + 1)(x² - x + 1) = 0

x² + x + 1 = 0
Δ < 0

x² - x + 1 = 0
Δ < 0

x₁ = 0
x₂ = 1
x₃ = -1

2) (x³ + 2x)(x³ + 2)(x³ + x) = 0
x²(x² + 2)(x³ + 2)(x² + 1) = 0
x²(x² + 2)(x + ³√2)(x² - x + ³√4)(x² + 1) = 0

x² + 2
Δ < 0

x² + 1
Δ < 0

x² - x + ³√4
Δ = 1 - 4³√4 < 1 - 4⁴√4 = 1 - 4√2 < 0

x₁ = 0
x₂ = - ³√2

3) (4x² - 8x + 6)(4x² - 8x)(- 8x + 6) = 0
2(2x² - 4x + 3)4x(x - 2)(-8)(x - 3/4) = 0
x(2x² - 4x + 3)(x - 2)(x - 3/4) = 0

2x² - 4x + 3 = 0
Δ = 16 - 24 < 0

x₁ = 0
x₂ = 2
x₃ = 3/4