ĐKXĐ:...
pt\(\Leftrightarrow4\left(x^2-2x\right)+16\sqrt{x^2-2x-3}-21=0\)
Đặt \(\sqrt{x^2-2x-3}=t\left(t\ge0\right)\Rightarrow t^2=x^2-2x-3\Leftrightarrow t^2+3=x^2-2x\)
\(\Rightarrow4\left(t^2+3\right)+16t-21=0\)
\(\Leftrightarrow4t^2+12+16t-21=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\frac{1}{2}\\t=-\frac{9}{2}\left(l\right)\end{matrix}\right.\Rightarrow t=\frac{1}{2}\)
\(\Rightarrow x^2-2x-3=\frac{1}{4}\Leftrightarrow\left[{}\begin{matrix}x=\frac{2+\sqrt{17}}{2}\\x=\frac{2-\sqrt{17}}{2}\left(l\right)\end{matrix}\right.\)
Vậy \(x=\frac{2+\sqrt{17}}{2}\)
