Đáp án: `(x;y;z)=(1/2; 2/5; 9/10)`.
Giải thích các bước giải:
`(2x-1)^{2008} + (y - 2/5)^{2008} + |x+y-z| = 0`
Vì `(2x-1)^{2008} + (y - 2/5)^{2008} + |x+y-z| ≥ 0 ∀ x;y;z`
`⇒ (2x-1)^{2008} + (y - 2/5)^{2008} + |x+y-z| = 0` khi :
$\left\{\begin{matrix}(2x-1)^{2008}=0& \\(y-\dfrac{2}{5})^{2008}=0&\\ |x+y-z|=0& \end{matrix}\right.$
$⇒$ $\left\{\begin{matrix}x = \dfrac{1}{2}&\\ y=\dfrac{2}{5}&\\x+y-z=0& \end{matrix}\right.$
$⇒$ $\left\{\begin{matrix}x = \dfrac{1}{2}&\\ y=\dfrac{2}{5}&\\x = x+y=\dfrac{1}{2} + \dfrac{2}{5} = \dfrac{9}{10}& \end{matrix}\right.$
Vậy `(x;y;z)=(1/2; 2/5; 9/10)`.
