Ta có:

$\left\{ \begin{align}& \left| x+\frac{17}{3} \right|\ge 0 \\ & \left| y-\frac{2000}{1999} \right|\ge 0 \\ &\left| z-2005 \right|\ge 0 \\ \end{align} \right.$  với mọi x, y, z nên suy ra:

$\left| x+\frac{17}{3} \right|+\left| y-\frac{2000}{1999} \right|+\left| z-2005 \right|=0$khi và chỉ khi

$\left\{ \begin{align}& \left| x+\frac{17}{3} \right|=0\Rightarrow x=-\frac{17}{3} \\ & \left| y-\frac{2000}{1999} \right|\ge 0\Rightarrow y=\frac{2000}{1999} \\ & \left| z-2005 \right|\ge 0\Rightarrow z=2005 \\ \end{align} \right.$

Vậy x + z = $-\frac{17}{3}+2005=\frac{5998}{3}$