\(A^2=\left(3\sqrt{x-1}+4\sqrt{5-x}\right)^2\le\left(3^2+4^2\right)\left(x-1+5-x\right)=25\cdot4=100\\ \Leftrightarrow A\le10\\ \Leftrightarrow A_{max}=10\Leftrightarrow\dfrac{x-1}{9}=\dfrac{5-x}{16}\Leftrightarrow x=\dfrac{61}{25}\\ A=3\left(\sqrt{x-1}+\sqrt{5-x}\right)+\sqrt{5-x}\ge3\sqrt{x-1+5-x}+\sqrt{5-x}\ge6+0=6\)
\(\Leftrightarrow A_{min}=6\Leftrightarrow x=5\)
Vậy \(6\le A\le10\)

