Question

\(\sqrt{x-2}-\sqrt{x+2}=2\sqrt{x^2-4}-2x+2\)

Answer 1

ĐKXĐ: \(x\ge2\)

\(\Leftrightarrow\left(2x-2\sqrt{x^2-4}\right)-\left(\sqrt{x+2}-\sqrt{x-2}\right)-2=0\)

Đặt \(\sqrt{x+2}-\sqrt{x-2}=t>0\Rightarrow t^2=2x-2\sqrt{x^2-4}\)

\(\Rightarrow t^2-t-2=0\Rightarrow\left[{}\begin{matrix}t=-1\left(l\right)\\t=2\end{matrix}\right.\)

\(\Rightarrow\sqrt{x+2}-\sqrt{x-2}=2\)

\(\Leftrightarrow\sqrt{x+2}=2+\sqrt{x-2}\)

\(\Leftrightarrow x+2=x+2+4\sqrt{x-2}\)

\(\Leftrightarrow x=2\)