Với mọi x, y, z ta có:

$\left\{ \begin{align}& {{(3x-5)}^{2006}}\ge 0 \\  & {{\left( {{y}^{2}}-1\right)}^{2010}}\ge 0 \\ & {{\left( x-z \right)}^{2016}}\ge 0 \\ \end{align}\right.$

Do đó: ${{\left( 3x-5 \right)}^{2006}}+{{({{y}^{2}}-1)}^{2010}}+{{(x-z)}^{2016}}=0$

$\Rightarrow \left\{ \begin{align}& {{(3x-5)}^{2006}}=0 \\ & {{\left( {{y}^{2}}-1 \right)}^{2010}}=0 \\ & {{\left( x-z \right)}^{2016}}=0 \\ \end{align} \right.$

$\Rightarrow \left\{ \begin{align}& x=\frac{5}{3} \\ & {{y}^{2}}=1 \\ & x=z \\\end{align} \right.$

Vậy:

A = |$x+{{y}^{2}}+z$ | = $\left| \frac{5}{3}+1+\frac{5}{3} \right|=\frac{13}{3}$