\(\sqrt{x-3}+\sqrt{5-x}=2x^2-7x-2\ (3\le x\le5)\) \(\\\) \(\Rightarrow(\sqrt{x-3}-1)+(\sqrt{5-x}-1)=2x^2-7x-4\)\(\\\) \(\Rightarrow\dfrac{x-4}{\sqrt{x-3}+1}+\dfrac{4-x}{\sqrt{5-x}+1}=(x-4)(2x+1)\)\(\\\) \(\Rightarrow\dfrac{x-4}{\sqrt{x-3}+1}-\dfrac{x-4}{\sqrt{5-x}+1}-(x-4)(2x+1)=0\)\(\\\) \(\Rightarrow(x-4)\left[\dfrac{1}{\sqrt{x-3}+1}-\dfrac{1}{\sqrt{5-x}+1}-(2x+1)\right]=0\)\(\\\) Trường hợp \(1:\) \(x-4=0\Rightarrow x=4\ (tm)\)\(\\\) Trường hợp \(2:\)\(\\\) \(\dfrac{1}{\sqrt{x-3}+1}-\dfrac{1}{\sqrt{5-x}+1}-(2x+1)=0\ (1)\) Với \(3\le x\le5\), ta có:\(\\\) \(\begin{cases}\sqrt{x-3}+1\ge 1\Rightarrow \dfrac{1}{\sqrt{x-3}+1}\le 1\\ \dfrac{1}{\sqrt{5-x}+1}>0\\ 2x+1\ge 7\end{cases}\)\(\\\) \(\Rightarrow\dfrac{1}{\sqrt{x-3}+1}-\dfrac{1}{\sqrt{5-x}+1}-(2x+1)<1-0-7=-6<0\) \(\\\) \(\Rightarrow \) Phương trình \((1)\) vô nghiệm. \(\\\) \(\Rightarrow\) Vậy \(x\in\{4\}\).