\(\sqrt{x-3}+\sqrt{5-x}-2x^2+7x+2=0\) (1)
ĐKXĐ: \(3\le x\le5\)
(1) \(\Leftrightarrow\sqrt{x-3}-1+\sqrt{5-x}-1-2x^2+8x-x+4=0\)
\(\Leftrightarrow\frac{x-4}{\sqrt{x-3}+1}+\frac{4-x}{\sqrt{5-x}+1}-2x\left(x-4\right)-\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(\frac{1}{\sqrt{x-3}+1}-\frac{1}{\sqrt{5-x}+1}-2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\\frac{1}{\sqrt{x-3}+1}-\frac{1}{\sqrt{5-x}+1}-2x-1=0\end{matrix}\right.\)
+) \(x-4=0\Leftrightarrow x=4\)( thỏa mãn)
+) \(\frac{1}{\sqrt{x-3}+1}-\frac{1}{\sqrt{5-x}+1}-2x-1=0\)
\(\Leftrightarrow\frac{1}{\sqrt{x-3}+1}=\frac{1}{\sqrt{5-x}+1}+2x+1\)
Ta có: \(VT=\frac{1}{\sqrt{x-3}+1}< 1\)
\(VP=\frac{1}{\sqrt{5-x}+1}+2x+1>\frac{1}{\sqrt{5-x}+1}+2.3+1>1\)
\(\Rightarrow\)vô nghiệm
Vậy phương trình có nghiệm là \(x=4\)
