Để\(\sqrt{x^2-4}+\sqrt{x-2}=0\) thì \(\left\{{}\begin{matrix}\sqrt{x^2-4}=0\\\sqrt{x-2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)\left(x+2\right)=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\x=2\end{matrix}\right.\Leftrightarrow x=2\)
+ Đk: 2 cái trong căn >/ 0 (tự xét ^^!)
+ pt đã cho <=> \(\left\{{}\begin{matrix}x^2-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=-2\left(L\right)\\x=2\left(N\right)\end{matrix}\right.\\x=2\left(N\right)\end{matrix}\right.\)
KL: x=2
