$\begin{array}{l} \dfrac{{\left( {x - 4} \right)\left( {{x^2} - 6x + 9} \right)\left( {{x^2} - 3x + 2} \right)}}{{\sqrt {2{x^2} + x + 2} - x}} \ge 0\\ \Leftrightarrow \dfrac{{\left( {x - 4} \right){{\left( {x - 3} \right)}^2}\left( {x - 1} \right)\left( {x - 2} \right)\left( {\sqrt {2{x^2} + x + 2} + x} \right)}}{{{x^2} + x + 2}} \ge 0\\ \Leftrightarrow \left( {x - 4} \right){\left( {x - 3} \right)^2}\left( {x - 1} \right)\left( {x - 2} \right)\left( {\sqrt {2{x^2} + x + 2} + x} \right) \ge 0\\ \left( {\text{do}\,{x^2} + x + 2 > 0} \right)\\ \left( {\text{lại có}\sqrt {2{x^2} + x + 2} + x > 0} \right)\\ \Leftrightarrow \left( {x - 4} \right){\left( {x - 3} \right)^2}\left( {x - 1} \right)\left( {x - 2} \right) \ge 0 \end{array}$
$\xrightarrow{{\,\,\,\,\,\,\,\,\,\, - \,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\, - \,\,\,\,\,\,\,3\,\,\,\,\,\,\,\,\, - \,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\,}}$
$\begin{array}{l} \Rightarrow x \in \left[ {1;2} \right] \cup \left[ {4; + \infty } \right) \cup \left\{ 3 \right\}\\ \Rightarrow x \in \left[ {1;2} \right] \cup \left[ {4;49} \right] \cup \left\{ 3 \right\}\left( {do\,x < 50} \right)\\ \Rightarrow \text{49 giá trị} \end{array}$
