Cho hàm số ${f\left( x \right) = 8{x^3} - 36{x^2} + 53x - 25 - m -...

Ta có: $8{{x}^{3}}-36{{x}^{2}}+53x-25-m-\sqrt[3]{3x-5+m}={{\left( 2x-3 \right)}^{3}}+2x-3-3x+5+m+\sqrt[3]{3x-5+m}\ge 0$
$\Leftrightarrow {{\left( 2x-3 \right)}^{3}}+2x-3\ge 3x-5+m+\sqrt[3]{3x-5+m}$
Gọi $f\left( x \right)={{\left( 2x-3 \right)}^{3}}+2x-3,g\left( x \right)=3x-5+m+\sqrt[3]{3x-5+m}\Leftrightarrow f\left( 2x-3 \right)\ge f\left( \sqrt[3]{3x-5+m} \right)$
Xét $f\left( t \right)={{t}^{3}}+t\Rightarrow f'\left( t \right)=2{{t}^{2}}+1>0,\forall t\in R$
Bất phương trình $\Leftrightarrow 2x-3\ge \sqrt[3]{3x-5+m}\Leftrightarrow {{\left( 2x-3 \right)}^{3}}\ge 3x-5+m\Leftrightarrow m\le {{\left( 2x-3 \right)}^{3}}-3x+5$
Gọi $h\left( x \right)={{\left( 2x-3 \right)}^{3}}-3x+5$
$\Rightarrow h'\left( x \right)=3{{\left( 2x-3 \right)}^{3}}-3=0\Leftrightarrow \left[ \begin{aligned}
& 2x-3=1 \\
& 2x-3=-1 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& x=2 \\
& x=1 \\
\end{aligned} \right.$
Ta có bàng biến thiên:

Từ đây ta suy ra: $m\le 0$ theo đề bài $-2019\le m\le 2019$ nên $2019\le m\le 0$ mà m là số nguyên |
Vậy có: 2019+1= 2020 số nguyên.