Question

Si:
$(x + y) {}^{2} = 4xy$
Calcula
$( \frac{x}{y} ) {}^{2020}$
​

Answer 1

Respuesta:

1

Explicación paso a paso:

${(x + y)}^{2} = 4xy$

${x}^{2} + 2xy + {y}^{2} = 4xy$

${x}^{2} + 2xy - 4xy + {y }^{2} = 0$

${x}^{2} - 2xy + {y}^{2} = 0$

${(x - y)}^{2} = 0$

$(x - y) = \sqrt{0}$

$x - y = 0$

$x = y$

$\frac{x}{y} = 1$

$[ \frac{x}{y} ]^{2020} = 1^{2020}$

$[ \frac{x}{y} ]^{2020} = 1$

Answer 2

Hola !! ^^

=> Resolviendo :

(x+y)^2 =4xy

x^2 + 2xy + y^2 = 4xy

x^2 + 2xy - 4xy + y^2 = 0

x^2 - 2xy + y^2 = 0

(x-y)^2 =0

x-y =0

x = y

=> Entonces :

(x/y)^2020

(x/x)^2020

1^2020

1

Respuesta : 1