Đặt \(f\left( x \right) = {x^5} + 2{x^4} - m{x^2} + 3x - 20 \Rightarrow y = \left| {f\left( x \right)} \right| \Rightarrow y' = \frac{{f'\left( x \right) \cdot f\left( x \right)}}{{\left| {f\left( x \right)} \right|}}\)

Yêu cầu bài toán \( \Leftrightarrow y' \le 0\,;\,\,\forall x \in \left( { - \infty \,;\,\, - 2} \right) \Leftrightarrow f'\left( x \right) \cdot f\left( x \right) \le 0\,;\,\,\forall x \in \left( { - \infty \,;\,\, - 2} \right).\)

• TH1: \(\left\{ {\begin{array}{*{20}{l}}{f'\left( x \right) \ge 0}\\{f\left( x \right) \le 0}\end{array};\forall x \in \left( { - \infty \,;\,\, - 2} \right) \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{5{x^4} + 8{x^3} - 2mx + 3 \ge 0}\\{f\left( { - 2} \right) \le 0}\end{array}} \right.} \right.\)

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{2mx \le 5{x^4} + 8{x^3} + 3;\forall x \in \left( { - \infty \,;\,\, - 2} \right)}\\{{{\left( { - 2} \right)}^5} + 2 \cdot {{\left( { - 2} \right)}^4} - m \cdot {{\left( { - 2} \right)}^2} + 3 \cdot \left( { - 2} \right) - 20 \le 0}\end{array}} \right.\)

\[ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{2m \ge \frac{{5{x^4} + 8{x^3} + 3}}{x};\,\,\forall x \in \left( { - \infty \,;\,\, - 2} \right)}\\{ - 4m - 26 \le 0}\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{2m \ge {{\max }_{\left( { - \infty \,;\,\, - 2} \right)}}\left( {5{x^3} + 8{x^2} + \frac{3}{x}} \right)}\\{ - 4m \le 26}\end{array}} \right.} \right.\]

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{2m \ge  - \frac{2}}\\{m \ge  - \frac{2}}\end{array} \Leftrightarrow m \ge  - \frac{4}} \right.\) mà \(m \in {\mathbb{Z}^ - }\) nên \[m \in \left\{ { - 4\,;\,\, - 3\,;\,\, - 2\,;\,\, - 1} \right\}.\]

• TH2: \(\left\{ {\begin{array}{*{20}{l}}{f'\left( x \right) \le 0}\\{f\left( x \right) \ge 0}\end{array};\,\,\forall x \in \left( { - \infty \,;\,\, - 2} \right) \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{5{x^4} + 8{x^3} - 2mx + 3 \le 0}\\{f\left( { - 2} \right) \ge 0}\end{array}} \right.} \right.\)

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{2mx \ge 5{x^4} + 8{x^3} + 3;\,\,\forall x \in \left( { - \infty \,;\,\, - 2} \right)}\\{{{\left( { - 2} \right)}^5} + 2 \cdot {{\left( { - 2} \right)}^4} - m \cdot {{\left( { - 2} \right)}^2} + 3 \cdot \left( { - 2} \right) - 20 \ge 0}\end{array}} \right.\)

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{2m \le \frac{{5{x^4} + 8{x^3} + 3}}{x};\,\,\forall x \in \left( { - \infty \,;\,\, - 2} \right)}\\{ - 4m - 26 \ge 0}\end{array}} \right.\)

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{2m \le {{\min }_{\left( { - \infty \,;\,\, - 2} \right)}}\left( {5{x^3} + 8{x^2} + \frac{3}{x}} \right)}\\{ - 4m \ge 26}\end{array}} \right.\)\( \Leftrightarrow m \in \emptyset .\)

Vậy có tất cả 4 giá trị nguyên của tham số \(m\) cần tìm.

Đáp án: 4.