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