Cho hàm số $y=f(x)=\left|x^4-4 x^2+4 m x+m+2017\right|$. Gọi $S$...

Đặt $g(x)=x^4-4 x^3+4 m x+m+2017 \Rightarrow y=f(x)=|g(x)|$. $g^{\prime}(x)=4 x^3-12 x^2+4 m$
Ta có hình vẽ minh họa
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Dạng bài toán này luôn chỉ xảy ra hai trường hợp $X$
$\begin{aligned}
& \left[ \begin{aligned}
& \left\{ \begin{aligned}
& g'\left( x \right)\ge 0,\forall x\in \left( -2;3 \right) \\
& g\left( -2 \right)\ge 0 \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& g'\left( x \right)\le 0,\forall x\in \left( -2;3 \right) \\
& g\left( -2 \right)\le 0 \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& \left\{ \begin{aligned}
& 4{{x}^{3}}-12{{x}^{2}}+4m\ge 0,\forall x\in \left( -2;3 \right) \\
& 2065-7m\ge 0 \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& 4{{x}^{3}}-12{{x}^{2}}+4m\le 0,\forall x\in \left( 2;3 \right) \\
& 2065-7m\le 0 \\
\end{aligned} \right. \\
\end{aligned} \right. \\
& \Leftrightarrow \left[ \begin{aligned}
& \left\{ \begin{aligned}
& m\ge -{{x}^{3}}+3{{x}^{2}},\forall x\in \left( -2;3 \right) \\
& 295\ge m \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& m\le -{{x}^{3}}+3{{x}^{2}},\forall x\in \left( -2;3 \right) \\
& m\ge 295 \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& \left\{ \begin{aligned}
& m\ge \underset{\left[ -2;3 \right]}{\mathop{\max }} \left( -{{x}^{3}}+3{{x}^{2}} \right)=20 \\
& m\le 295 \\
\end{aligned} \right. \\
& \left\{ \begin{aligned}
& m\ge \underset{\left[ -2;3 \right]}{\mathop{\min }} \left( -{{x}^{3}}+3{{x}^{2}} \right)=0 \\
& m\ge 295 \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow 20\le m\le 295 \\
\end{aligned}$
Suy ra có 276 giá trị $m$ nguyên thỏa mãn.