Giải thích các bước giải:
Ta có:
$\begin{array}{l}
y = {x^3} + m{x^2} - x + m\\
\Rightarrow y' = 3{x^2} + 2mx - 1
\end{array}$
Để $y' < 0,\forall x \in \left( {1;2} \right)$
$\begin{array}{l}
\Leftrightarrow 3{x^2} + 2mx - 1 < 0,\forall x \in \left( {1;2} \right)\\
\Leftrightarrow 2mx < 1 - 3{x^2},\forall x \in \left( {1;2} \right)\\
\Leftrightarrow m < \dfrac{1}{{2x}} - \dfrac{{3x}}{2},\forall x \in \left( {1;2} \right)\\
\Leftrightarrow m < \mathop {\min }\limits_{x \in \left( {1;2} \right)} \left( {\dfrac{1}{{2x}} - \dfrac{{3x}}{2}} \right)\left( 1 \right)
\end{array}$
Đặt $f\left( x \right) = \dfrac{1}{{2x}} - \dfrac{{3x}}{2},x \in \left( {1;2} \right)$
$\begin{array}{l}
\Rightarrow f'\left( x \right) = \dfrac{{ - 1}}{{2{x^2}}} - \dfrac{3}{2} < 0,\forall x \in \left( {1;2} \right)\\
\Rightarrow \mathop {\min }\limits_{x \in \left( {1;2} \right)} f\left( x \right) = f\left( 2 \right) = \dfrac{{ - 11}}{4}\left( 2 \right)
\end{array}$
Từ $\left( 1 \right),\left( 2 \right) \Rightarrow m < \dfrac{{ - 11}}{4}$
Vậy $m < \dfrac{{ - 11}}{4}$ thỏa mãn đề.
