Решите плиз [tex] \frac{x}{x-2} + \frac{8}{4- x^{2} } - \frac{1}{x+2} =0[/tex]. Created by VADIM1234565432. algebra-ru

ОДЗ:(x = 2 не уд. ОДЗ).Ответ: x = -3.

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

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Dимасuk

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

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