Đáp án:
\(m \geq 8\)
Giải thích các bước giải:
\[\begin{array}{l}
4\sqrt {\left( {x + 1} \right)\left( {3 - x} \right)} \le {x^2} - 2x + m - 3\\
\Leftrightarrow 4\sqrt {3 + 2x - {x^2}} - {x^2} + 2x + 3 \le m\\
\Leftrightarrow m \ge \left( { - {x^2} + 2x + 3} \right) + 4\sqrt {3 + 2x - {x^2}} \,\,\,\left( * \right)\\
Dat\,\,\,\sqrt {3 + 2x - {x^2}} = t\,\, \Rightarrow {t^2} = 3 + 2x - {x^2}\\
x \in \left[ { - 1;\,\,3} \right] \Rightarrow t \in \left[ {0;\,\,2} \right]\\
\Rightarrow \left( * \right) \Leftrightarrow m \ge {t^2} + 4t\\
\Leftrightarrow m \ge \mathop {Max}\limits_{\left[ {0;\,\,2} \right]} \left( {{t^2} + 4t} \right)\\
Co\,\,\,f\left( t \right) = {t^2} + 4t \Rightarrow f'\left( t \right) = 2t + 4 = 0 \Leftrightarrow t = - 2\,\, \notin \left[ {0;\,\,2} \right]\\
f\left( 0 \right) = 0;\,\,\,\,f\left( 2 \right) = 8\\
\Rightarrow \mathop {Max}\limits_{\left[ {0;\,\,2} \right]} \left( {{t^2} + 4t} \right) = 8\\
\Rightarrow m \ge 8.
\end{array}\]
