(LONDRINA) Dados os números $\phantom{X}x\,=\,\dfrac{\dfrac{1}{3}\,+\,\dfrac{1}{3}}{\dfrac{1}{3}}\phantom{X}$, $\phantom{X}y\,=\,\dfrac{\dfrac{1}{3}\,+\,\dfrac{1}{3}}{\dfrac{3}{2}}\phantom{X}$, $\phantom{X}z\,=\,\dfrac{\dfrac{\dfrac{1}{3}\,+\,\dfrac{1}{3}}{3}}{\dfrac{1}{2}}\phantom{X}$ pode-se concluir que:
