\(\sqrt{x-2}+\sqrt{4-x}=2x^2-5x-1\left(2\le x\le4\right)\)
\(\Leftrightarrow2x^2-5x-1-\sqrt{x-2}-\sqrt{4-x}=0\)
\(\Leftrightarrow2x^2-5x-3-\left(\sqrt{x-2}-1\right)-\left(\sqrt{4-x}-1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-3\right)-\dfrac{x-3}{\sqrt{x-2}+1}+\dfrac{x-3}{\sqrt{4-x}+1}=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x-1-\dfrac{1}{\sqrt{x-2}+1}+\dfrac{1}{\sqrt{4-x}+1}\right)=0\left(2\right)\)
Vì \(2\le x\le4\)
\(\Leftrightarrow3-1+\dfrac{1}{\sqrt{2}+1}\le2x-1-\dfrac{1}{\sqrt{x-2}+1}+\dfrac{1}{\sqrt{4-x}+1}\le7-\dfrac{1}{\sqrt{2}+1}+1\)
\(\Leftrightarrow2+\dfrac{1}{\sqrt{2}+1}\le2x-1-\dfrac{1}{\sqrt{x-2}+1}+\dfrac{1}{\sqrt{4-x}+1}\le8-\dfrac{1}{\sqrt{2}+1}\)
\(\Rightarrow2x-1-\dfrac{1}{\sqrt{x-2}+1}+\dfrac{1}{\sqrt{4-x}+1}>0\)
\(\left(2\right)\Rightarrow x-3=0\)
\(\Leftrightarrow x=3\left(tm\right)\)
Vậy nghiệm của phương trình cho có 1 nghiệm duy nhất là \(x\in\left\{3\right\}\)