Đáp án:
$\begin{array}{l}
a)f\left( x \right) = \sqrt {\left( {5 - x} \right)\left( {x + 3} \right)} \\
= \sqrt { - {x^2} + 5x - 3x + 15} \\
= \sqrt { - {x^2} + 2x + 15} \\
= \sqrt { - {x^2} + 2x - 1 + 16} \\
= \sqrt { - {{\left( {x - 1} \right)}^2} + 16} \\
Do: - {\left( {x - 1} \right)^2} \le 0\forall - 3 \le x \le 5\\
\Rightarrow - {\left( {x - 1} \right)^2} + 16 \le 16\forall - 3 \le x \le 5\\
\Rightarrow f\left( x \right) \le \sqrt {16} = 4\forall - 3 \le x \le 5\\
\Rightarrow GTLN:f\left( x \right) = 4 \Leftrightarrow x = 1\\
b)f\left( x \right) = \sqrt {{x^2}\left( {6 - x} \right)} \left( {0 \le x \le 6} \right)\\
Do:\left\{ \begin{array}{l}
0 \le {x^2} \le 36\,\forall 0 \le x \le 6\\
0 \le 6 - x \le 6\forall 0 \le x \le 6
\end{array} \right.\\
\Rightarrow {x^2}\left( {6 - x} \right) \le 32\\
\Rightarrow \sqrt {{x^2}\left( {6 - x} \right)} \le 4\sqrt 2 \\
\Rightarrow GTLN:f\left( x \right) = 4\sqrt 2 \Leftrightarrow x = 4
\end{array}$
