ĐK: \(0\le x\le3\)
(Mà VT phương trình không âm nên điều kiện kéo theo \(x^2-x-2\ge0\))
\(PT\Leftrightarrow\sqrt{x}-\left(x-1\right)+\sqrt{3-x}-\left(x-2\right)+2x-3=x^2-x-2\)
\(\Leftrightarrow\frac{x-x^2+2x-1}{\sqrt{x}+x-1}+\frac{3-x-x^2+4x-4}{\sqrt{3-x}+x-2}-x^2+3x-1=0\)
\(\Leftrightarrow\left(-x^2+3x-1\right)\left(\frac{1}{\sqrt{x}+x-1}+\frac{1}{\sqrt{3-x}+x-2}+1\right)=0\)
Với \(0\le x\le3\) \(\Rightarrow\left\{{}\begin{matrix}x+\sqrt{x}-1\le\sqrt{3}+2\\x+\sqrt{3-x}-2\le\sqrt{3}+1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\frac{1}{x+\sqrt{x}-1}>0\\\frac{1}{x+\sqrt{3-x}-2}>0\end{matrix}\right.\)
\(\Rightarrow\frac{1}{x+\sqrt{x}-1}+\frac{1}{x+\sqrt{3-x}-2}+1=0\) vô no
\(\Leftrightarrow-x^2+3x-1=0\Leftrightarrow x=\frac{3\pm\sqrt{5}}{2}\)
Thử lại thấy \(x=\frac{3+\sqrt{5}}{2}\) thỏa mãn.
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