Có tất cả bao nhiêu giá trị nguyên của tham số $m$ thuộc đoạn...

Xét $f\left( x \right)=-{{x}^{3}}+3\left( m+1 \right){{x}^{2}}-3m\left( m+2 \right)x+{{m}^{2}}\left( m+3 \right)$
${f}'\left( x \right)=-3{{x}^{2}}+6\left( m+1 \right)x-3m\left( m+2 \right)$
${f}'\left( x \right)=0\Leftrightarrow -3{{x}^{2}}+6\left( m+1 \right)x-3m\left( m+2 \right)=0$ $\Leftrightarrow \left[ \begin{aligned}
& x=m \\
& x=m+2 \\
\end{aligned} \right.$.
Hàm số $y=\left| -{{x}^{3}}+3\left( m+1 \right){{x}^{2}}-3m\left( m+2 \right)x+{{m}^{2}}\left( m+3 \right) \right|$ đồng biến trên khoảng $\left( 0,1 \right)$
$\Leftrightarrow $ $\left[ \begin{aligned}
& \left\{ \begin{aligned}
& f\left( x \right)\text{ db }\forall x\in \left( 0,1 \right) \\
& f\left( 0 \right)\ge 0 \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& f\left( x \right)\text{ nb }\forall x\in \left( 0,1 \right) \\
& f\left( 0 \right)\le 0 \\
\end{aligned} \right. \\
\end{aligned} \right.$$\Leftrightarrow $ $ \left[ \begin{aligned}
& \left\{ \begin{aligned}
& {f}'\left( x \right)\ge 0\text{ }\forall x\in \left( 0,1 \right)\text{ }\left( I \right) \\
& f\left( 0 \right)\ge 0 \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& {f}'\left( x \right)\le 0\text{ }\forall x\in \left( 0,1 \right)\text{ }\left( II \right) \\
& f\left( 0 \right)\le 0 \\
\end{aligned} \right. \\
\end{aligned} \right.$
Giải $\left( I \right)\Leftrightarrow \left\{ \begin{aligned}
& -3{{x}^{2}}+6\left( m+1 \right)x-3m\left( m+2 \right)\ge 0\text{ }\forall x\in \left( 0,1 \right) \\
& f\left( 0 \right)\ge 0 \\
\end{aligned} \right. $ $ \Leftrightarrow \left\{ \begin{aligned}
& m\le 0<1\le m+2 \\
& {{m}^{2}}\left( m+3 \right)\ge 0 \\
\end{aligned} \right. $ $ \Leftrightarrow \left\{ \begin{aligned}
& -1\le m\le 0 \\
& m\ge -3 \\
\end{aligned} \right.\Leftrightarrow -1\le m\le 0$.
Giải $\left( II \right)\Leftrightarrow \left\{ \begin{aligned}
& -3{{x}^{2}}+6\left( m+1 \right)x-3m\left( m+2 \right)\le 0\text{ }\forall x\in \left( 0,1 \right) \\
& f\left( 0 \right)\le 0 \\
\end{aligned} \right. $ $ \Leftrightarrow \left\{ \begin{aligned}
& \left[ \begin{aligned}
& m+2\le 0 \\
& m\ge 1 \\
\end{aligned} \right. \\
& {{m}^{2}}\left( m+3 \right)\le 0 \\
\end{aligned} \right. $ $ \Leftrightarrow \left\{ \begin{aligned}
& \left[ \begin{aligned}
& m\le -2 \\
& m\ge 1 \\
\end{aligned} \right. \\
& m\le -3 \\
\end{aligned} \right.\Leftrightarrow m\le -3$.
Kết hợp $\left( I \right)$ và $\left( II \right)$ suy ra $\left[ \begin{aligned}
& -1\le m\le 0 \\
& m\le -3 \\
\end{aligned} \right.$$\xrightarrow[m\in \mathbb{Z}]{m\in \left[ -2023,2023 \right]}m\in \left\{ -2023,-2022,....-3,-1,0 \right\}$.
Vậy có $2023$ giá trị $m$ thỏa mãn đề bài.