Question

tìm điều kiện để biểu thức xác định

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

Answer 1

\(\text{ĐKXĐ: }x^2-4\ge0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2\ge0\\x+2\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2\le0\\x+2\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2\\x\ge-2\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2\\x\le-2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge2\\x\le-2\end{matrix}\right.\)

\(\RightarrowĐKXĐ:x\ge2;x\le-2\)