Ответ:

[tex]-2\frac{1}{6};\ 2[/tex]

Объяснение:

[tex]\frac{x^2-6y^2}{xy}=-1\\\\x^2-6y^2=-xy\\\\x^2-6y^2+xy=0\\\\x^2-9y^2+xy+3y^2=0\\\\(x+3y)(x-3y)+y(x+3y)=0\\\\(x+3y)(x-3y+y)=0\\\\(x+3y)(x-2y)=0\\\\x+3y=0\ \ \ \ \ \ \ \ \ \ x-2y-0\\\\x=-3y\ \ \ \ \ \ \ \ \ \ \ \ \ \ x=2y[/tex]

[tex]x=-3y\\\\\frac{x^2+4y^2}{2xy}=\frac{(-3y)^2+4y^2}{2(-3y)y}=\frac{9y^2+4y^2}{-6y^2}=\frac{13y^2}{-6y^2}=-\frac{13}{6}=-2\frac{1}{6}[/tex]

[tex]x=2y\\\\\frac{x^2+4y^2}{2xy}=\frac{(2y)^2+4y^2}{2\cdot 2y}=\frac{4y^2+4y^2}{4y^2}=\frac{8y^2}{4y^2}=2[/tex]