Question

[DRACULA]

giải pt: \(4\left(x^2-2x\right)+16\sqrt{\left(x-3\right)\left(x+1\right)}-21=0\)

Answer 1

[Hoàng Tử Hà]

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 \(x=\frac{2+\sqrt{17}}{2}\)