Помогите по алгебре.

1. Разложу на множители знаменатель:
ax² + bx + c = a(x - x₁)(x - x₂)
x₁ и x₂ - корни уравнения ax² + bx + c = 0

6x² + 11x - 2 = 0
D = 121 - 4*(-2)*6 = 169
√D = 13
x₁ = (-11-13)/12 = -2
x₂ = (-11+13)/12 = 1/6

6x² + 11x - 2 = 6(x - 1/6)(x + 2) = (6x - 1)(x + 2)

$\frac{x^3+8}{6x^2 + 11x - 2} = \frac{(x+2)(x^2-2x+4)}{ (6x - 1)(x + 2)} = \frac{x^2-2x+4}{ 6x - 1}$

2. -x² + 5x - 6 = 0
D = 25 - 4*(-1)*(-6) = 1
√D = 1
x₁ = (-5-1)/(-2) = 3
x₂ = (-5+1)/(-2) = 2

-x² + 5x - 6 = -(x - 3)(x - 2) = (3 - x)(x - 2)

$\frac{x^2-4}{5x -x^2- 6} = \frac{(x-2)(x+2)}{(3-x)(x-2)} = \frac{x+2}{3-x}$

3. -x² + 4x - 3 = 0
D = 16 - 4*(-3)*(-1) = 4
√D = 2
x₁ = (-4-2)/(-2) = 3
x₂ = (-4+2)/(-2) = 1

-x² + 4x - 3 = -(x - 1)(x - 3) = (1 - x)(x - 3)

x² + 2x - 15 = 0
D = 4 - 4*(-15) = 64
√D = 8
x₁ = (-2-8)/2 = -5
x₂ = (-2+8)/2 = 3

x² + 2x - 15 = (x - 3)(x + 5)

$\frac{x^2 + 2x - 15}{4x-x^2 - 3} = \frac{(x - 3)(x + 5)}{(1 - x)(x - 3)} = \frac{x + 5}{1 - x}$