ĐK: \(x\le-1;5\le x\)
Đặt \(x^2-4x-5=a\), pt trở thành:
\(2x^2-8x-13+\sqrt{a}=0\\ \Leftrightarrow2\left(x^2-4x-5\right)-3+\sqrt{a}=0\\ \Leftrightarrow2a+\sqrt{a}-3=0\\ \Leftrightarrow\left(\sqrt{a}-1\right)\left(2\sqrt{a}+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{a}=1\\\sqrt{a}=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=1\\a\in\varnothing\left(\sqrt{a}\ge0\right)\end{matrix}\right.\\ \Leftrightarrow x^2-4x-5=1\\ \Leftrightarrow x^2-4x-6=0\\ \Delta=16+24=40\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4-\sqrt{40}}{2}\\x=\dfrac{4+\sqrt{40}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2\left(2-\sqrt{10}\right)}{2}=2-\sqrt{10}\left(tm\right)\\x=\dfrac{2\left(2+\sqrt{10}\right)}{2}=2+\sqrt{10}\left(tm\right)\end{matrix}\right.\)
