Vì \(\left|x-\frac{2}{3}\right|\ge0\); \(\left|2y+3\right|\ge0\); \(\left(z-2\right)^2\ge0\)
=> \(\left|x-\frac{2}{3}\right|+\left|2y+3\right|+\left(z-2\right)^2\ge0\)
Mà theo đề bài: \(\left|x-\frac{2}{3}\right|+\left|2y+3\right|+\left(z-2\right)^2=0\)
=> \(\begin{cases}\left|x-\frac{2}{3}\right|=0\\\left|2y+3\right|=0\\\left(z-2\right)^2=0\end{cases}\)=> \(\begin{cases}x-\frac{2}{3}=0\\2y+3=0\\z-2=0\end{cases}\)=> \(\begin{cases}x=\frac{2}{3}\\2y=-3\\z=2\end{cases}\)=> \(\begin{cases}x=\frac{2}{3}\\y=-\frac{3}{2}\\z=2\end{cases}\)
Vậy \(x=\frac{2}{3};y=-\frac{3}{2};z=2\)