Điều kiện: \(x^2-3x+2\ge0\)\(\Leftrightarrow x\ge2\)
\(\sqrt{x^2-3x+2}=x^2-3x-4\)
\(\Leftrightarrow x^2-3x+2-2\cdot\dfrac{1}{2}\sqrt{x^2-3x+2}+\dfrac{1}{4}=\dfrac{25}{4}\)
\(\Leftrightarrow\left(\sqrt{x^2-3x+2}-\dfrac{1}{2}\right)^2=\dfrac{25}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-3x+2}-\dfrac{1}{2}=\dfrac{5}{2}\\\sqrt{x^2-3x+2}-\dfrac{1}{2}=\dfrac{-5}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-3x+2}=3\\\sqrt{x^2-3x+2}=-2\end{matrix}\right.\)
Đối chiếu điều kiện ta được \(\sqrt{x^2-3x+2}=3\)
\(\Leftrightarrow x^2-3x+2=9\)
\(\Leftrightarrow x^2-3x-7=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt{37}}{2}\\x=\dfrac{3-\sqrt{37}}{2}\end{matrix}\right.\)
Đối chiếu điều kiện \(\Rightarrow x=\dfrac{3+\sqrt{37}}{2}\)
