Đáp án:
$\begin{array}{l}
2{x^2} - 4x + 5 + m \ge 0\forall x \in \left[ { - 1;4} \right]\\
\Rightarrow 2{x^2} - 4x + 5 \ge - m\forall x \in \left[ { - 1;4} \right]\\
Xet:f\left( x \right) = 2{x^2} - 4x + 5\\
\Rightarrow \mathop {\min }\limits_{\left[ { - 1;4} \right]} f\left( x \right) \ge m\\
f\left( x \right) = 2\left( {{x^2} - 2x + 1} \right) + 3\\
= 2{\left( {x - 1} \right)^2} + 3 \ge 3\forall x \in \left[ { - 1;4} \right]\\
\Rightarrow \min \,f\left( x \right) = 3\\
\Rightarrow m \le 3\\
Do:m \in \left( { - 20;20} \right)\\
\Rightarrow m \in \left\{ { - 20; - 19;...;1;2;3} \right\}
\end{array}$
Vậy có 24 giá trị nguyên của m
