Question

Срочно! Даю 100 баллов

Answer 1

Ответ: 1

$\displaystyle\\\frac{x}{y} =\frac{1}{2}~~~\Rightarrow~~~y=2x\\\\\\\frac{4x^2-y^2+2xy}{3y^2-8x^2} =\frac{4x^2-(2x)^2+2x\cdot(2x)}{3\cdot(2x)^2-8x^2} =\\\\\\=\frac{4x^2-4x^2+4x^2}{12x^2-8x^2} =\frac{4x^2}{4x^2}=1$