Question

Se
$xy = 12 \\ x ^{2} + y ^{2} = 24$

então
$( \frac{1}{x} + \frac{1}{y} )^{2}$
vale:

A:
$\frac{1}{3}$
B:
$\frac{1}{2}$
C: 1

D:2

E:3
​

Answer 1

$\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2 = \dfrac{1}{x^2}+2\cdot\dfrac{1}{xy}+\dfrac{1}{y^2} \\~\\\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2 = \dfrac{x^2+y^2}{(xy)^2}+2\cdot \dfrac{1}{xy}\\~\\\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2 = \dfrac{1}{xy}\cdot\left(\dfrac{x^2+y^2}{xy}+2\right)\\~\\\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2 =\dfrac{1}{12}\cdot\left(\dfrac{24}{12}+2\right)\\~\\\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2 =\dfrac{4}{12}\\~\\\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2 = \dfrac{1}{3}$