Question

Даю 50б Решите уравнение:

Answer 1

$\dfrac{x2}{x^2+2x}-\dfrac{3}{x^2+2x-2}=1\ \ ,\ \ \ ODZ:\ \left\{\begin{array}{l}x^2+2x\ne 0\\x^2+2x-2\ne 0\end{array}\right\\\\\\t=x^2+2x\ \ ,\ \ \ \dfrac{12}{t}-\dfrac{3}{t-2}=1\ \ ,\ \ \dfrac{12(t-2)-3t-t(t-2)}{t(t-2)}=0\ ,\\\\\\\dfrac{12t-24-3t-t^2+2t}{t(t-2)}=0\ \ ,\ \ \ \dfrac{-t^2+11t-24}{t(t-2)}=0\ ,\\\\\\t^2-11t+24=0\ \ ,\ \ t_1=3\ ,\ \ t_2=8\ \ (teorena\ Vieta)\\\\a)\ \ x^2+2x=3\ \ ,\ \ x^2+2x-3=0\ \ ,\ \ x_1=-3\ ,\ x_2=1\\\\b)\ \ x^2+2x=8\ \ ,\ \ x^2+2x-8=0\ \ ,x_3=-4\ ,\ x_4=2$

$x_1,\ x_2,\ x_3,\ x_4\in ODZ\\\\Otvet:\ \ x_1=-3,\ x_2=1,\ x_3=-4,\ x_4=2\ .$