\(x\ge\dfrac{31}{8}\)
\(9x-19=6\sqrt{8x-31}+4\sqrt{x-1}\)
\(\Leftrightarrow8x-31-6\sqrt{8x-31}+9+x-1-4\sqrt{x-1}+4=0\)
\(\Leftrightarrow\left(\sqrt{8x-31}-3\right)^2+\left(\sqrt{x-1}-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{8x-31}-3=0\\\sqrt{x-1}-2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}8x-31=9\\x-1=4\end{matrix}\right.\) \(\Rightarrow x=5\)
Vậy pt có nghiệm duy nhất \(x=5\)