Question

Найдите сумму всех корней уравнения (x−1/x^3+3*x^2+x+3)+(1/x^4−1)=(x+1/x^3+3*x^2−x−3).

Answer 1

$\frac{x-1}{x^3+3*x^2+x+3}+ \frac{1}{x^4-1}= \frac{x+1}{x^3+3*x^2-x-3}$
$\frac{x-1}{x(x^2+1)+3(x^2+1)}+ \frac{1}{(x^2-1)(x^2+1)}= \frac{x+1}{x(x^2-1)+3(*x^2+1)}$
$\frac{x-1}{(x+3)(x^2+1)}+ \frac{1}{(x^2-1)(x^2+1)}= \frac{x+1}{(x+3)(x^2-1)}$
$\frac{(x-1)(x^2-1)+x+3-(x+1)(x^2+1)}{(x+3)(x^2+1)(x^2-1)}=0$
$\frac{x^3-x-x^2+1+x+3-x^3-x-x^2-1}{(x+3)(x^2+1)(x^2-1)}=0$
$\frac{-x-2x^2+3}{(x+3)(x^2+1)(x^2-1)}=0$
$\frac{-2(x-1)(x-1,5)}{(x+3)(x^2+1)(x^2-1)}=0$
$x_1=1$∈ОДЗ
$x_2=1,5$
ответ: 1,5