Đáp án:
$\begin{array}{l}
A = 3\sqrt {x - 1} + 4\sqrt {5 - x} \\
Dkxd:1 \le x \le 5\\
Theo\,Bunhia:\\
{A^2} = {\left( {3\sqrt {x - 1} + 4\sqrt {5 - x} } \right)^2}\\
\le \left( {{3^2} + {4^2}} \right).\left( {x - 1 + 5 - x} \right)\\
\Rightarrow {A^2} \le 25.4 = 100\\
\Rightarrow - 10 \le A \le 10\\
Do:A > 0\\
\Rightarrow 0 < A \le 10\\
\Rightarrow GTLN:A = 10\,khi:\dfrac{{\sqrt {x - 1} }}{3} = \dfrac{{\sqrt {5 - x} }}{4}\\
\Rightarrow 4\sqrt {x - 1} = 3\sqrt {5 - x} \\
\Rightarrow 16\left( {x - 1} \right) = 9\left( {5 - x} \right)\\
\Rightarrow 16x - 16 = 45 - 9x\\
\Rightarrow 25x = 61\\
\Rightarrow x = \dfrac{{61}}{{25}}\left( {tmdk} \right)
\end{array}$
