Ответ:

[tex]\displaystyle 1[/tex]

Объяснение:

[tex]\displaystyle x > 0,\ y > 0\\\\\frac{x^2-4y^2}{xy}=-3\\\\x^2-4y^2=-3xy\\\\x^2-4y^2+3xy=0\\\\x^2-y^2-3y^2+3xy=0\\\\(x-y)(x+y)+3y(-y+x)=0\\\\(x-y)(x+y)+3y(x-y)=0\\\\(x-y)(x+y+3y)=0\\\\(x-y)(x+4y)=0\\\\x-y=0\ \ \ \ \ \ \ \ \ \ x+4y=0\\\\x=y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x=-4y[/tex]

але x>o i y>0 ось чому

[tex]\displaystyle x=y[/tex]

[tex]\displaystyle \frac{2x^2+y^2}{3xy}=\frac{2y^2+y^2}{3y\cdot y}=\frac{3y^2}{3y^2}=1[/tex]