ĐKXĐ: \(x\ge2\)
Do \(\sqrt{x+2}+\sqrt{x-2}>0\), nhân 2 vế của pt với \(\sqrt{x+2}+\sqrt{x-2}\)
\(\Leftrightarrow4\left(1+\sqrt{x^2-4}\right)=4\left(\sqrt{x+2}+\sqrt{x-2}\right)\)
\(\Leftrightarrow\sqrt{x^2-4}-\sqrt{x+2}-\left(\sqrt{x-2}-1\right)=0\)
\(\Leftrightarrow\sqrt{x+2}\left(\sqrt{x-2}-1\right)-\left(\sqrt{x-2}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x+2}-1\right)\left(\sqrt{x-2}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+2}-1=0\\\sqrt{x-2}-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-1\left(l\right)\\x=3\end{matrix}\right.\)