Question

помогите пожалуйста я дам 22 баллов​

Answer 1

Ответ:

$\left\{\begin{array}{l}\dfrac{x}{y}-\dfrac{y}{x}=\dfrac{3}{2}\\x^2-y^2=3\end{array}\right\ \ \left\{\begin{array}{l}\dfrac{x^2-y^2}{xy}=\dfrac{3}{2}\\x^2-y^2=3\end{array}\right\ \ \left\{\begin{array}{l}\dfrac{3}{xy}=\dfrac{3}{2}\\x^2-y^2=3\end{array}\right\ \ \left\{\begin{array}{l}xy=2\\x^2-y^2=3\end{array}\right$

$\left\{\begin{array}{l}y=\dfrac{2}{x}\\x^2-\dfrac{4}{x^2}=3\end{array}\right\ \ \left\{\begin{array}{l}\ y=\dfrac{2}{x}\\\dfrac{x^4-4}{x^2}=3\end{array}\right\ \ \left\{\begin{array}{l}\ y=\dfrac{2}{x}\\x^4-3x^2-4=0\end{array}\right\\\\\\x^4-3x^2-4=0\ \ ,\ \ t=x^2\geq 0\ \ ,\ \ \ t^2-3t-4=0\ ,\ t_1=-1\ ,\ t_2=4\\\\x^2=4\ \ ,\ \ x=\pm 2\\\\a)\ \ x_1=-2\ ,\ y_1=\dfrac{2}{-2}=-1\\\\b)\ \ x_2=2\ \ ,\ \ y_2=\dfrac{2}{2}=1\\\\Otvet:\ \ (-2\ ;-1\ )\ ,\ (\ 2\ ;\ 1\ )\ .$