Ответ:

[tex](-1;-3);\ (1;3)[/tex]

Объяснение:

[tex]x,y \neq 0[/tex]

[tex]\begin{cases}x^2-y^2=-8\\xy=3\end{cases}[/tex]

[tex]\begin{cases}x^2-y^2=-8\\y=\frac{3}{x}\end{cases}[/tex]

[tex]\begin{cases}x^2- \left(\frac{3}{x}\right)^2=-8\\y=\frac{3}{x}\end{cases}[/tex]

[tex]x^2- \left(\frac{3}{x}\right)^2=-8[/tex]

[tex]x^2-\frac{9}{x^2}=-8[/tex]

[tex]\frac{x^4}{x^2}-\frac{9}{x^2}+\frac{8x^2}{x^2}=0[/tex]

[tex]\frac{x^4-9+8x^2}{x^2}=0[/tex]

[tex]x^4+8x^2-9=0[/tex]

подставляем

[tex]x^2=t, t > 0\\\\t^2+8t-9=0\\\\t^2+9t-t-9=0\\\\t(t+9)-(t+9)=0\\\\(t+9)(t-1)=0\\\\t+9=0\ \ \ \ \ \ \ \ \ \ \ \ \ \ t-1=0\\\\t=-9 < 0\ \ \ \ \ \ \ \ \ \ \ \ t=1[/tex]

[tex]x^2=1\\\\x^2-1=0\\\\(x-1)(x+1)=0\\\\x-1=0\ \ \ \ \ \ \ \ \ \ x+1=0\\\\x=1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=-1[/tex]

[tex]\begin{cases}x=1\\y=\frac{3}{x}\end{cases}\\\\\begin{cases}x=1\\y=\frac{3}{1}\end{cases}\\\\\begin{cases}x=1\\y=3\end{cases}[/tex]                                           [tex]\begin{cases}x=-1\\y=\frac{3}{x}\end{cases}\\\\\begin{cases}x=-1\\y=\frac{3}{-1}\end{cases}\\\\\begin{cases}x=-1\\y=-3\end{cases}[/tex]