می‌دانیم $\sqrt {{x^2}}  = \left| x \right|$ و $\left| x \right| = \left\{ \begin{gathered}
  x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \geqslant 0 \hfill \\
   - x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \lt 0 \hfill \\ 
\end{gathered}  \right.$ پس عبارت را ساده می‌کنیم.

$\sqrt {{x^2}}  + \left( {\sqrt {{{(5 - x)}^2}} } \right)\left( {\sqrt {{{(x - 7)}^2}} } \right) + \left| {2x - 8} \right| = \left| x \right| + \left| {5 - x} \right|\left| {x - 7} \right| + \left| {2x - 8} \right|$

چون $ - 3 \lt x \lt  - 1$ پس: $x \lt 0 \to \left| x \right| =  - x$

$\begin{gathered}
  \left| {5 - x} \right| \to  - 3 \lt x \lt  - 1\xrightarrow{{ \times ( - 1)}}3 \gt  - x \gt 1\xrightarrow{{ + 5}}8 \gt 5 - x \gt 6 \hfill \\
   \Rightarrow 5x \gt 0 \to \left| {5 - x} \right| = 5 - x \hfill \\ 
\end{gathered} $

$\left| {x - 7} \right| \to  - 3 \lt x \lt  - 1\xrightarrow{{ + ( - 7)}} - 10 \lt x - 7 \lt  - 8 \to x - 7 \lt 0 \to \left| {x - 7} \right| =  - (x - 7)$

$\begin{gathered}
  \left| {2x - 8} \right| \to  - 3 \lt x \lt  - 1\xrightarrow{{ \times 2}} - 6 \lt 2x \lt  - 2 \hfill \\
  \xrightarrow{{ + ( - 8)}} - 14 \lt 2x - 8 \lt  - 10 \to 2x - 8 \lt 0 \to \left| {2x - 8} \right| =  - (2x - 8) \hfill \\ 
\end{gathered} $

$\begin{gathered}
   \Rightarrow \left| x \right| + \left| {5 - x} \right|\left| {x - 7} \right| + \left| {2x - 8} \right| \hfill \\
   =  - x + \underbrace {\left( {5 - x} \right)\left( { - \left( {x - 7} \right)} \right)}_{{x^2} - 12x + 35} - \left( {2x - 8} \right) \hfill \\
   =  - x + {x^2} - 12x + 35 - 2x + 8 = {x^2} - 15x + 43 \hfill \\ 
\end{gathered} $