Помогите решить уравнение:
[tex]\frac{6}{(x+1)(x+2)} + \frac{8}{(x-1)(x+4)} = 1 [/tex]

ОДЗ

x ≠ - 1

x ≠ - 2

x ≠ 1

x ≠ - 4

6(x - 1)(x + 4) + 8(x + 1)(x + 2) - (x + 1)(x + 2)(x - 1)(x + 4) = 0

6x^2 + 18x - 24 + 8x^2 + 24x + 16 - x^4 - 6x^3 - 7x^2 + 6x + 8 = 0

- x^4 - 6x^3 + 7x^2 + 48x = 0

x^4 + 6x^3 - 7x^2 - 48x = 0

x (x^3 + 6x^2 - 7x - 48) = 0

x (x^3 + 3x^2 + 3x^2 + 9x - 16x - 48) = 0

x ( x^2 (x + 3) + 3x(x +3) - 16(x + 3)) = 0

x (x + 3)(x^2 + 3x - 16) = 0

1) x = 0

2) x + 3 = 0

x = - 3

3) x^2 + 3x - 16 = 0

D = 73

x1 = ( - 3 - √73)/2

x2 = ( - 3 + √73)/2

ОТВЕТ

- 3; 0 ; ( - 3 - √73)/2; ( - 3 + √73)/2