Cho hàm số ${f\left( x \right)=2{{e}^{-x}}-\log {{\left(...

Điều kiện, $m\sqrt{{{x}^{2}}+1}-mx>0\Rightarrow m<0.$
Khi đó $f\left( x \right)=2{{e}^{x}}-log{{\left( m\sqrt{{{x}^{2}}+1}-mx \right)}^{3}}=2{{e}^{-x}}-3log\left( \sqrt{{{x}^{2}}+1-x} \right)-3logm$
$\Rightarrow f\left( -x \right)=2.{{e}^{x}}-\text{3log }\left( \sqrt{{{x}^{2}}+1}+x \right)-3logm=2.{{e}^{x}}3log\dfrac{1}{\sqrt{{{x}^{2}}+1}-x}-3logm$
$=2.{{e}^{x}}+3log\left( \sqrt{{{x}^{2}}+1}-x \right)-3logm$
Xét $f\left( x \right)+f\left( -x \right)=\left[ 2{{e}^{-x}}3log\left( \sqrt{{{x}^{2}}+1}-x \right)3logm \right]+\left[ 2.{{e}^{x}}+3log\left( \sqrt{{{x}^{2}}+1}x \right)3logm \right]$
$=2\left( {{e}^{x}}+{{e}^{-x}} \right)-6logm.$
Ta có $f\left( x \right)+f(-x)\ge 0\forall x\in \mathbb{R}\Leftrightarrow 2({{e}^{x}}+{{e}^{-x}})-6logm\ge 0\forall x\in \mathbb{R}$
$\Leftrightarrow 3\log m\le {{e}^{x}}+{{e}^{-x}}\forall x\in \mathbb{R}\left( * \right)$
Xét $g\left( x \right)={{e}^{x}}+{{e}^{-x}}={{e}^{x}}+\dfrac{1}{{{e}^{x}}}\ge 2,$ dấu bằng xảy ra khi $x=0,$ suy ra ming $g\left( x \right)=2.$
Khi đó $\left( * \right)3logm\le mi{{n}_{\mathbb{R}}}g\left( x \right)\Leftrightarrow 3logm\le 2\Leftrightarrow m\le \sqrt[3]{100}$.
Kết hợp điều kiện, ta được $m\in \left( 0;\sqrt[3]{100} \right).$
Do $m\in \mathbb{Z}\Rightarrow m\in \left\{ 1;2;3;4 \right\},$ gồm 4 giá trị.