Question

Система уравнений..............​

Answer 1

$\left\{\begin{array}{l}xy^3+x^3y=-10\\\\x^2y^4+x^4y^2=20\end{array}\right\ \ \left\{\begin{array}{l}xy\, (y^2+x^2)=-10\\\\x^2y^2\, (y^2+x^2)=20\end{array}\right\ \ \left\{\begin{array}{l}y^2+x^2=-\dfrac{10}{xy}\ ,\ \ xy\ne 0\\y^2+x^2=\dfrac{20}{(xy)^2}\ ,\ \ xy\ne 0\end{array}\right$

$-\dfrac{10}{xy}=\dfrac{20}{(xy)^2}\ \ \Rightarrow \ \ \ -10\cdot (xy)^2=20\cdot (xy)\ \ ,\ \ \ (xy)^2=-2\cdot (xy)\ \ ,\\\\\\(xy)^2+2\cdot (xy)=0\ \ \ ,\ \ \ xy\cdot (xy+2)=0\ \ \Rightarrow \ \ xy=0\ \ ili\ \ xy=-2\\\\\\y^2+x^2=-\dfrac{10}{-2}=5$

$\left\{\begin{array}{l}xy=-2\\\\x^2+y^2=5\end{array}\right\ \ \left\{\begin{array}{l}y=-\dfrac{2}{x}\\x^2+\dfrac{4}{x^2}-5=0\end{array}\right\ \ \left\{\begin{array}{l}y=-\dfrac{2}{x}\\\\x^4-5x^2+4=0\ ,\ x\ne 0\end{array}\right$

$x^4-5x^2+4=0\ \ \ \Rightarrow \ \ \ x^2=1\ \ \ ili\ \ \ x^2=4\\\\x_1=-1\ ,\ x_2=1\ \ \ ili\ \ \ x_3=-2\ ,\ x_4=2\\\\y_1=-\dfrac{2}{x_1}=-\dfrac{2}{-1}=2\ \ ,\ \ \ y_2=-\dfrac{2}{x_2}=-\dfrac{2}{1}=-2\\\\\\y_3=-\dfrac{2}{x_3}=-\dfrac{2}{-3}=1\ \ ,\ \ y_4=-\dfrac{2}{x_4}=-\dfrac{2}{2}=-1\\\\\\Otvet:\ \ (-1\, ;\, 2\, )\ ,\ (\, 1\, ;-2\, )\ ,\ (-2\, ;\, 1)\ ,\ (\, 2\, ;\, -1\, )\ .$