Yanıt: $\boxed{A}$
$x=0$ ise $(x,y,z)=(0,0,0)$ dır.
$x \neq 0$ ise
$\dfrac{1}{x} = \dfrac{1}{4z^2} + 1$
$\dfrac{1}{y} = \dfrac{1}{4x^2} + 1$
$\dfrac{1}{z} = \dfrac{1}{4y^2} + 1$
$\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = \dfrac{1}{4x^2} + 1 + \dfrac{1}{4y^2} + 1 + \dfrac{1}{4z^2} + 1$
$0 = \left (\dfrac{1}{2x}-1 \right )^2 + \left (\dfrac{1}{2y}-1 \right )^2 + \left (\dfrac{1}{2z}-1 \right )^2$
$2x = 2y = 2z = 1 \Rightarrow (x,y,z) = \left (\dfrac 12, \dfrac 12, \dfrac 12 \right )$.