Question

Алгебра 9 класс. Срочно нужна помощь!!!​

Answer 1

одз х≠-1, х≠1, х≠0

$\frac{6 - x}{(1 - x)(1 + x)} - \frac{x + 3}{x(1 - x)} - \frac{x + 5}{x(1 + x)} = 0 \\ \frac{x (6 - x) - (1 + x)(x + 3) - (1 - x)(x + 5)}{x(1 - x)(1 + x)} = 0 \\ \frac{6 - {x}^{2} - (x + 3 + {x}^{2} + 3x) - (x + 5 - {x}^{2} - 5x) }{x(1 - x)(1 + x)} = 0 \\ \frac{6x - 4x - 3 - {x}^{2} + 4 - 5}{x(1 - x)(1 + x)} = 0 \\ \frac{6x - 8 - {x}^{2} }{x(1 - x)(1 + x)} = 0 \\ 6x - 8 - {x}^{2} = 0 \\ {x}^{2} - 6x + 8 = 0 \\ x(x - 2) - 4(x - 2) = 0 \\ (x - 2)(x - 4) = 0 \\ x - 2 = 0 \\ x - 4 = 0 \\ \\ x = 2 \\ x = 4$