Ta có: $\left| \left| 3x-3 \right|+2x+{{\left( -1 \right)}^{2016}} \right|=3x+{{2017}^{0}}$
$\Rightarrow \left| \left| 3x-3 \right|+2x+1 \right|=3x+1$ (*)
Điều kiện có nghiệm của phương trình: $3x+1 \ge 0\Rightarrow x \ge \frac{-1}{3}$
Trường hợp 1: $3x-3 < 0\Rightarrow x < 1$
Khi đó: (*) $\Rightarrow \left| -3x+3+2x+1 \right|=3x+1$
$\Rightarrow \left| 4-x \right|=3x+1$
$\Rightarrow \left[ \begin{align}
& 4-x=3x+1 \\
& x-4=3x+1 \\
\end{align} \right.$$\Rightarrow \left[ \begin{align}
& 4x=3 \\
& 2x=-5 \\
\end{align} \right.$$\Rightarrow \left[ \begin{align}
& x=\frac{3}{4}\,(tm) \\
& x=\frac{-5}{2}\,(ktm) \\
\end{align} \right.$
Trường hợp 2: $3x-3 \ge 0\Rightarrow x \ge 1$
Khi đó (*) $\Rightarrow \left| 3x-3+2x+1 \right|=3x+1$
$\Rightarrow $$\left| 5x-2 \right|=3x+1$
$\Rightarrow \left[ \begin{align}
& 5x-2=3x+1 \\
& 2-5x=3x+1 \\
\end{align} \right.$$\Rightarrow \left[ \begin{align}
& 2x=3 \\
& 8x=1 \\
\end{align} \right.$$\Rightarrow \left[ \begin{align}
& x=\frac{3}{2}\,(tm) \\
& x=\frac{1}{8}\,(ktm) \\
\end{align} \right.$
Vậy $x\in \left\{ \frac{3}{4};\frac{3}{2} \right\}$