$A=2{{x}^{5}}-8{{x}^{4}}-5{{x}^{3}}+31{{x}^{2}}-19x+1$
$=\left( 2{{x}^{5}}-8{{x}^{4}}+2{{x}^{3}} \right)-\left( 7{{x}^{3}}-28{{x}^{2}}+7x \right)+\left( 3{{x}^{2}}-12x+3 \right)-2$$=2{{x}^{3}}\left( {{x}^{2}}-4x+1 \right)-7x\left( {{x}^{2}}-4x+1 \right)+3\left( {{x}^{2}}-4x+1 \right)-2$
Thay $x=2+\sqrt{3}$vào ta được ${{x}^{2}}-4x+1=0$nên ta có:
$A=2{{x}^{3}}.0-7x.0+3.0-2$ = -2
Vậy tại $x=2+\sqrt{3}$thì A = -2

$\begin{align}& \,\,\,\,\,2{{x}^{5}}-8{{x}^{4}}-5{{x}^{3}}+31{{x}^{2}}-19x+1 \\ & ={{x}^{4}}\left( 2x-8 \right)-5{{x}^{3}}+28{{x}^{2}}+3{{x}^{2}}-19x+1 \\ & ={{(2+\sqrt{3})}^{4}}.\left[ 2\left( 2+\sqrt{3} \right)-8 \right]-5{{\left( 2+\sqrt{3} \right)}^{3}}+28{{\left( 2+\sqrt{3} \right)}^{2}}+3{{\left( 2+\sqrt{3} \right)}^{2}}-19\left( 2+\sqrt{3} \right)+1 \\ & ={{(2+\sqrt{3})}^{4}}.\left( 2\sqrt{3}-4 \right)-5{{\left( 2+\sqrt{3} \right)}^{3}}+28{{\left( 2+\sqrt{3} \right)}^{2}}+3{{\left( 2+\sqrt{3} \right)}^{2}}-19\left( 2+\sqrt{3} \right)+1 \\ & ={{(2+\sqrt{3})}^{4}}.2\left( \sqrt{3}-2 \right)-5{{\left( 2+\sqrt{3} \right)}^{3}}+28{{\left( 2+\sqrt{3} \right)}^{2}}+3{{\left( 2+\sqrt{3} \right)}^{2}}-19\left( 2+\sqrt{3} \right)+1 \\ & ={{(2+\sqrt{3})}^{3}}.(-2)-5{{\left( 2+\sqrt{3} \right)}^{3}}+28{{\left( 2+\sqrt{3} \right)}^{2}}+3{{\left( 2+\sqrt{3} \right)}^{2}}-19\left( 2+\sqrt{3} \right)+1 \\ & =-7{{\left( 2+\sqrt{3} \right)}^{3}}+28{{\left( 2+\sqrt{3} \right)}^{2}}+3{{\left( 2+\sqrt{3} \right)}^{2}}-19\left( 2+\sqrt{3} \right)+1 \\ & =7.{{\left( 2+\sqrt{3} \right)}^{2}}.\left[ -2-\sqrt{3}+4 \right]+3{{\left( 2+\sqrt{3} \right)}^{2}}-19\left( 2+\sqrt{3} \right)+1 \\ & =7.{{\left( 2+\sqrt{3} \right)}^{2}}.\left( 2-\sqrt{3} \right)+3{{\left( 2+\sqrt{3} \right)}^{2}}-19\left( 2+\sqrt{3} \right)+1 \\ & =7\left( 2+\sqrt{3} \right)+3{{\left( 2+\sqrt{3} \right)}^{2}}-19\left( 2+\sqrt{3} \right)+1 \\ & =3{{\left( 2+\sqrt{3} \right)}^{2}}-12\left( 2+\sqrt{3} \right)+1 \\ & =3\left( 7+4\sqrt{3} \right)-12\left( 2+\sqrt{3} \right)+1 \\ & =21+12\sqrt{3}-24-12\sqrt{3}+1 \\ & =-2 \\ \end{align}$