đây là cách tôi nghĩ là ổn. mời bạn tham khảo

Hàm số nghịch biến trên [-1;2]
<=> y' = 0 AA x \in [-1;2]
<=> x^{3} - 9x^{2} + m - 2024 \le 0, AA x \in [-1;2]
<=> m \le - x^{3} + 9x^{2} + 2024, AA x \in [-1;2]
<=> m \le min (- x^{3} + 9x^{2} + 2024), AA x \in [-1;2]
Xét f (x) = - x^{3} + 9x^{2} + 2024 có:
f'(x) = - 3x^{2} + 18x
Ta có: f'(x) = -3x^{2} + 18x = 0 <=> [(x = 6),(x = 0):}
BBT: Hàm số f(x) = -x^{3} + 9x^{2} + 2024
$$\color{#FA8072}{\text{___}}$$ $$\color{#F28D84}{\text{___}}$$ $$\color{#EFAE98}{\text{___}}$$ $$\color{#E7C2AB}{\text{___}}$$ $$\color{#DFD6BF}{\text{___}}$$ $$\color{#D7E9D3}{\text{___}}$$ $$\color{#CFFDD7}{\text{___}}$$ $$\color{#C7FBEA}{\text{___}}$$ $$\color{#BFF9FE}{\text{___}}$$ $$\color{#B7F8F7}{\text{___}}$$ $$\color{#AFF6EC}{\text{___}}$$ $$\color{#A7F5DF}{\text{___}}$$ $$\color{#9FF3D2}{\text{___}}$$ $$\color{#97F2C5}{\text{___}}$$ $$\color{#8FF1B9}{\text{___}}$$ $$\color{#87F0AC}{\text{___}}$$ $$\color{#7FEFA0}{\text{___}}$$ $$\color{#77ED93}{\text{___}}$$
(Hình vẽ)
$$\color{#FA8072}{\text{___}}$$ $$\color{#F28D84}{\text{___}}$$ $$\color{#EFAE98}{\text{___}}$$ $$\color{#E7C2AB}{\text{___}}$$ $$\color{#DFD6BF}{\text{___}}$$ $$\color{#D7E9D3}{\text{___}}$$ $$\color{#CFFDD7}{\text{___}}$$ $$\color{#C7FBEA}{\text{___}}$$ $$\color{#BFF9FE}{\text{___}}$$ $$\color{#B7F8F7}{\text{___}}$$ $$\color{#AFF6EC}{\text{___}}$$ $$\color{#A7F5DF}{\text{___}}$$ $$\color{#9FF3D2}{\text{___}}$$ $$\color{#97F2C5}{\text{___}}$$ $$\color{#8FF1B9}{\text{___}}$$ $$\color{#87F0AC}{\text{___}}$$ $$\color{#7FEFA0}{\text{___}}$$ $$\color{#77ED93}{\text{___}}$$
=> m \le 2024
Kết hợp điều kiện {(m \in [6;26]),(m \in ZZ):}, ta có: m = {6; 7; ... ; 26}
Vậy có 21 giá trị m thỏa mãn
-> B. 21
$$\color{#FA8072}{\text{$$\textit{$$\circ$$ hungnguyen4269}$$}}$$