Решите плиз [latex] \frac{x}{x-2} + \frac{8}{4- x^{2} } - \frac{1}{x+2} =0[/latex]

ОДЗ: [latex]x \neq \pm 2 \\ \\ \frac{x}{x-2} + \frac{8}{4-x^2} - \frac{1}{x+2} = 0 \\ \\ \frac{x}{x-2} + \frac{8}{(2-x)(2 +x)} - \frac{1}{x+2} = 0 \\ \\ \frac{x}{x-2} - \frac{8}{(x-2)(2 +x)} - \frac{1}{x+2} = 0 \\ \\ \frac{x(x+2)}{(x-2)(x+2)} - \frac{8}{(x-2)(2 +x)} - \frac{x-2}{(x+2)(x+2)} = 0 \\ \\ \frac{x^2 + 2x - x + 2 - 8}{x^2 - 4} = 0 \\ \\ x^2 + x - 6 = 0 \\ x_1 + x_2 = -1 \\ x_1*x_2 = -6 \\ \\ x_1 = -3 \\ x_2 = 2 [/latex] (x = 2 не уд. ОДЗ). Ответ: x = -3.