Đáp án:

\(\begin{array}{l}
1)\,a = \dfrac{{51}}{4}\\
2)\,GTNN\,la\,\dfrac{7}{4}
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
P\left( x \right) = \dfrac{1}{2}\left( {2{x^4} - 10{x^2} + 8x} \right) + a\\
 = \dfrac{1}{2}\left[ {2{x^4} + {x^3} - {x^3} - \dfrac{1}{2}{x^2} - \dfrac{{19}}{2}{x^2} - \dfrac{{19}}{4}x + \dfrac{{51}}{4}x + \dfrac{{51}}{8}} \right] + a - \dfrac{{51}}{4}\\
 = \dfrac{1}{2}\left[ {{x^3}\left( {2x + 1} \right) - \dfrac{1}{2}{x^2}\left( {2x + 1} \right) - \dfrac{{19}}{4}x\left( {2x + 1} \right) + \dfrac{{51}}{8}\left( {2x + 1} \right)} \right] + a - \dfrac{{51}}{4}\\
 = \dfrac{1}{2}\left( {2x + 1} \right)\left( {{x^3} - \dfrac{1}{2}{x^2} - \dfrac{{19}}{4}x + \dfrac{{51}}{8}} \right) + a - \dfrac{{51}}{4}\\
De\,P\left( x \right)\, \vdots \,Q\left( x \right)\,thi\,a - \dfrac{{51}}{4} = 0 \Leftrightarrow x = \dfrac{{51}}{4}\\
2)\,{x^2} + x + 2 = {\left( {x + \dfrac{1}{2}} \right)^2} + \dfrac{7}{4} \ge \dfrac{7}{4}\\
 \Rightarrow GTNN\,la\,\dfrac{7}{4} \Leftrightarrow x =  - \dfrac{1}{2}
\end{array}\)