ĐKXĐ: \(1\le x\le5\)
\(A=3\sqrt{x-1}+4\sqrt{5-x}>0\)
\(A^2=9\left(x-1\right)+16\left(5-x\right)+24\sqrt{\left(x-1\right)\left(5-x\right)}\)
\(A^2=9\left(x-1\right)+9\left(5-x\right)+7\left(5-x\right)+24\sqrt{\left(x-1\right)\left(5-x\right)}\)
\(A^2=9\left(x-1+5-x\right)+7\left(5-x\right)+24\sqrt{\left(x-1\right)\left(5-x\right)}\)
\(A^2=36+7\left(5-x\right)+24\sqrt{\left(x-1\right)\left(5-x\right)}\)
Do \(x\le5\Rightarrow\left\{{}\begin{matrix}7\left(5-x\right)\ge0\\24\sqrt{\left(x-1\right)\left(5-x\right)}\ge0\end{matrix}\right.\)
\(\Rightarrow A^2\ge36\Rightarrow A\ge6\)
\(A_{min}=6\) khi \(x=5\)
