Vì :
$\bigg(x-\dfrac{1}{5}\bigg)^{2004} ≥ 0 ∀x$
$(y+0,4)^{100} ≥0 ∀y$
$(z-3)^{678} ≥0 ∀z $
Do đó :
$\bigg(x-\dfrac{1}{5}\bigg)^{2004} + (y+0,4)^{100} + (z-3)^{678} ≥0∀x,y,z $
Mà theo đề ta có : $\bigg(x-\dfrac{1}{5}\bigg)^{2004} + (y+0,4)^{100} + (z-3)^{678} =0 $
Nên : $ \left\{ \begin{array}{l}(x-\dfrac{1}{5}\bigg)^{2004}=0\\(y+0,4)^{100}=0\\(z-3)^{678} = 0\end{array} \right.$
$⇔ \left\{ \begin{array}{l}(x=\dfrac{1}{5}\\y=-0,4\\z=3\end{array} \right.$
Vậy $(x,y,z)=\bigg(\dfrac{1}{5};-0,4;3\bigg)$
