Question

найдите сумму всех корней уравнения

Answer 1

$\dfrac{x-1}{x^{3} + 3x^{2} + x + 3} + \dfrac{1}{x^{4} - 1} = \dfrac{x+1}{x^{3} + 3x^{2} - x - 3}$

$\dfrac{x-1}{x^{2}(x+3) + (x+3)} + \dfrac{1}{(x^{2}-1)(x^{2} + 1)} = \dfrac{x+1}{x^{2}(x+3)-(x+3)}$

$\dfrac{x-1}{(x+3)(x^{2}+1)} + \dfrac{1}{(x-1)(x+1)(x^{2} + 1)} = \dfrac{x+1}{(x+3)(x-1)(x+1)}$

$\dfrac{x-1}{(x+3)(x^{2}+1)} + \dfrac{1}{(x-1)(x+1)(x^{2} + 1)} - \dfrac{x+1}{(x+3)(x-1)(x+1)} = 0$

ОДЗ:

$1) \ (x+3)(x^{2}+1) \neq 0 \ \ \ \ \ \ \ \ \ \\ 2) \ (x-1)(x+1)(x^{2}+1)\neq 0 \\ 3) \ (x+3)(x-1)(x+1)\neq 0$

$x \neq -3; \ x\neq \pm1$

$\dfrac{x-1}{(x+3)(x^{2}+1)} + \dfrac{1}{(x-1)(x+1)(x^{2} + 1)} - \dfrac{x+1}{(x+3)(x-1)(x+1)} = 0$

$\dfrac{x-1}{(x+3)(x^{2}+1)} + \dfrac{1}{(x-1)(x+1)(x^{2} + 1)} - \dfrac{1}{(x+3)(x-1)} = 0$

$\dfrac{(x+1)(x-1)^{2} + x + 3 - (x+1)(x^{2}+1)}{(x+1)(x-1)(x^{2}+1)(x+3)} = 0$

$(x+1)(x-1)^{2} + x + 3 - (x+1)(x^{2}+1) = 0\\(x+1)(x^{2}-2x+1) + x + 3 - (x^{3} + x + x^{2} + 1)=0 \\x^{3} - 2x^{2} + x + x^{2} - 2x + 1 + x + 3 - x^{3} - x - x^{2} - 1 = 0\\-2x^{2} - x + 3 = 0 \ \ \ | : (-2)\\x^{2} + 0,5x - 1,5 = 0\\x_{1} + x_{2} =-0,5$

Ответ: -0,5.