Giải thích

Xét \(u(x) = \frac{{{x^4} + ax + a}}{{x + 1}}\) trên đoạn [1; 2], ta có \({u^\prime }(x) = \frac{{3{x^4} + 4{x^3}}}{{{{(x + 1)}^2}}} > 0,\forall x \in [1;2]\).

Do đó, \({\max _{[1;2]}}u = u(2) = a + \frac{{16}}{3},{\min _{[1;2]}}u = u(1) = a + \frac{1}{2}\).

TH1. Nếu \(a + \frac{1}{2} \ge 0 \Rightarrow \left\{ {\begin{array}{*{20}{l}}{M = a + \frac{{16}}{3}}\\{m = a + \frac{1}{2}}\end{array} \Rightarrow \left\{ {\begin{array}{*{20}{l}}{a + \frac{1}{2} \ge 0}\\{a + \frac{{16}}{3} \ge 2\left( {a + \frac{1}{2}} \right)}\end{array} \Leftrightarrow  - \frac{1}{2} \le a \le \frac{{13}}{3}} \right.} \right.\).

TH2. Nếu \(a + \frac{{16}}{3} \le 0 \Rightarrow \left\{ {\begin{array}{*{20}{l}}{M =  - \left( {a + \frac{1}{2}} \right)}\\{m =  - \left( {a + \frac{{16}}{3}} \right)}\end{array} \Rightarrow \left\{ {\begin{array}{*{20}{l}}{a + \frac{{16}}{3} \le 0}\\{ - \left( {a + \frac{1}{2}} \right) \ge  - 2\left( {a + \frac{{16}}{3}} \right)}\end{array} \Leftrightarrow  - \frac{{61}}{6} \le a \le  - \frac{{16}}{3}} \right.} \right.\).

TH3. Nếu \(\left( {a + \frac{1}{2}} \right).\left( {a + \frac{{16}}{3}} \right) \le 0 \Leftrightarrow \frac{{ - 16}}{3} \le a \le \frac{{ - 1}}{2} \Rightarrow m = 0\),

\(M = \max \left\{ {\left| {a + \frac{1}{2}} \right|,\left| {a + \frac{{16}}{3}} \right|} \right\} \Rightarrow M > 2m\) (thỏa mãn).

Vậy \( - \frac{{61}}{6} \le a \le \frac{{13}}{3}\). Mà \(a \in \mathbb{Z}\) nên \(a \in \{  - 10; \ldots ;4\} \).