ĐK : \(2\le x\le4\)
pt <=> \(\sqrt{x-2}+\sqrt{4-x}-\left(2x^2-5x+1\right)=0\)
\(\Leftrightarrow\sqrt{x-2}-1+\sqrt{4-x}-1-\left(2x^2-5x+3\right)=0\)
\(\Leftrightarrow\frac{x-3}{\sqrt{x-2}+1}+\frac{3-x}{\sqrt{4-x}+1}-\left(x-3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[\frac{1}{\sqrt{x-2}+1}-\frac{1}{\sqrt{4-x}+1}-\left(2x+1\right)\right]=0\)
TH1 : x - 3 = 0 <=> x = 3 ( tmđk )
TH2 : \(\frac{1}{\sqrt{x-2}+1}-\frac{1}{\sqrt{4-x}+1}-\left(2x+1\right)=0\)( tự xử lý nhe == , vô nghiệm á )
Vậy pt có nghiệm duy nhất là x = 3
