Chọn B.

Ta có

$3^{-\left|x^3-\frac{x^2}{2}-m\right|}{\log }_{\sqrt[3]{3}}\left(x^2-2x+5\right)+3^{-x^2+2x}{\log }_{\frac{1}{3}}\left(\left|x^3-\frac{x^2}{2}-m\right|+4\right)=0$

$\iff \frac{3}{3^{\left|x^3-\frac{x^2}{2}-m\right|}}.{\log }_3\left(x^2-2x+5\right)={\log }_3\left(\left|x^3-\frac{x^2}{2}-m\right|+4\right){.3}^{-x^2+2x}$

$\iff {\log }_3\left(x^2-2x+1+4\right){.3}^{x^2-2x+1}={\log }_3\left(\left|x^3-\frac{x^2}{2}-m\right|+4\right){.3}^{\left|x^3-\frac{x^2}{2}-m\right|},\left(*\right)$

Xét hàm số $f\left(t\right)=3^t.{\log }_3\left(t+4\right),\forall t\ge 0$

Ta có $f\text{'}\left(t\right)=3^t.\ln 3.{\log }_3\left(t+4\right)+\frac{1}{\left(t+4\right).\ln 3}>0,\forall t\ge 0.$

Suy ra hàm số f(t) đồng biến trên $\left(0;+\infty \right)$

Như vậy $\left(*\right)\Rightarrow x^2-2x+1=\left|x^3-\frac{x^2}{2}-m\right|\iff \left[\begin{array}{l}m=x^3-\frac{3}{2}{x}^2+2x-1,\left(1\right)\\ m=x^3+\frac{1}{2}{x}^2-2x+1,\left(2\right)\end{array}\right.,\left(**\right)$

Đặt $\left\{\begin{array}{l}h\left(x\right)=x^3-\frac{3}{2}{x}^2+2x-1\\ g\left(x\right)=x^3+\frac{1}{2}{x}^2-2x+1\end{array}\right.$