ĐKXĐ: \(2\le x\le4\)
pt\(\Leftrightarrow\sqrt{x-2}-\sqrt{4-x}=2x^2-5x-3\)
\(\Leftrightarrow\dfrac{2x-6}{\sqrt{x-2}+\sqrt{4-x}}=\left(2x+1\right)\left(x-3\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=\dfrac{2}{\sqrt{x-2}+\sqrt{4-x}}\left(1\right)\end{matrix}\right.\)
Xét (1), ta có: \(2x+1\ge5;\forall x\ge2\)
\(\sqrt{x-2}+\sqrt{4-x}\ge\sqrt{x-2+4-x}=\sqrt{2}\)
\(\Rightarrow\dfrac{2}{\sqrt{x-2}+\sqrt{4-x}}\le\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
\(\Rightarrow2x+1>\dfrac{2}{\sqrt{x-2}+\sqrt{4-x}}\)
\(\Rightarrow\left(1\right)\) vô nghiệm
Vậy pt có nghiệm duy nhất \(x=3\)
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