Question

Ненулевые действительные числа х,у удовлетворяют равенству 2х^2+2у^2=5ху. Найти х+у/х-у

Answer 1

1) 2x² + 2y² = 5xy

2[(x² + 2xy + y²) - 2xy] = 5xy

2[(x + y)²- 2xy] = 5xy

2(x + y)² - 4xy = 5xy

2(x + y)² = 9xy

(x + y)² = 4,5xy

2) 2[(x² - 2xy + y²) + 2xy] = 5xy

2[(x - y)²+ 2xy] = 5xy

2(x - y)² + 4xy = 5xy

2(x - y)² = xy

(x-+ y)² = 0,5xy

$\frac{(x+y)^{2} }{(x-y)^{2}}=\frac{4,5xy}{0,5xy}\\\\(\frac{x+y}{x-y})^{2}=9\\\\\left[\begin{array}{ccc}\frac{x+y}{x-y}=-3 \\\frac{x+y}{x-y} =3\end{array}\right$

Answer 2

Объяснение:

$2x^2+2y^2=5xy\\(\sqrt{2}*x)^2+(\sqrt{2}*y)^2=5xy\\\left \{ {{(\sqrt{2}*x)^2+2*\sqrt{2}*x*\sqrt{2}*y+(\sqrt{2}*y)^2=5xy+2*\sqrt{2}*x*\sqrt{2}y } \atop {(\sqrt{2}*x)^2-2*\sqrt{2}*x*\sqrt{2}*y+(\sqrt{2}*y)^2=5xy-2*\sqrt{2}*x*\sqrt{2}y }} \right. \\\left \{ {{(\sqrt{2}*x)^2+4xy+(\sqrt{2}*y)^2=5xy+4xy } \atop {(\sqrt{2}*x)^2+4xy+(\sqrt{2}*y)^2=5xy+4xy }} \right. \ \ \ \ \left \{ {{(\sqrt{2}*x+\sqrt{2}*y)^2=9xy } \atop {(\sqrt{2}*x-\sqrt{2}*y)^2=xy}} \right.$

$\left \{ {{2*(x+y)^2=9xy} \atop {2*(x-y)^2=xy}} \right. .$

Разделим первое уравнение на второе:

$\frac{(x+y)^2}{(x-y)^2}=9\\(\frac{x+y}{x-y})^2=9\\\frac{x+y}{x-y}=3,\ \ \ \ \ \frac{x+y}{x-y}=-3.$