Cho hàm số f(x) = (m^3 - 1)x^3 + 3x^2 + 3(m - 2)x + 4

Ta có: $f\left(x\right)\le 0$  với $\forall x\in \left[3;5\right].$

$\iff \left(m^3-1\right)x^3+3x^2+3\left(m-2\right)x+4\le \mathrm{0,}\forall x\in \left[3;5\right].$

$\iff {\left(mx\right)}^3+3mx\le x^3-3x^2+6x-\mathrm{4,}\forall x\in \left[3;5\right].$

$\iff {\left(mx\right)}^3+3mx\le {\left(x-1\right)}^3+3\left(x-1\right),\forall x\in \left[3;5\right].$

 $\iff g\left(mx\right)\le g\left(x-1\right)$ với $g\left(t\right)=t^3+3t$   là hàm số đồng biến.

$\iff mx\le x-\mathrm{1,}\forall x\in \left[3;5\right]\iff m\le \frac{x-1}{x}=1-\frac{1}{x}=h\left(x\right),\forall x\in \left[3;5\right]\iff m\le \underset{\left[3;5\right]}{\min }h\left(x\right).$

Ta có $h^{\text{'}}\left(x\right)=\frac{1}{x^2}>\mathrm{0,}\forall x\in \left[3;5\right],$ suy ra $h\left(x\right)$  đồng biến trên  $\left[3;5\right]\Rightarrow \underset{\left[3;5\right]}{\min }h\left(x\right)=h\left(3\right)=\frac{2}{3}.$

Vậy $m\le \frac{2}{3}\underset{m\in \mathbb{Z}}{\overset{m\in \left[-100;100\right]}{\to }}m:-100\to \mathrm{0,}$  nghĩa là có 101 số nguyên m.