lầm bài này nhé
a) \(\sqrt{x^2-4x-12}\le x-4\)
\(ĐK:\left\{{}\begin{matrix}\left[{}\begin{matrix}x\le-2\\x\ge6\end{matrix}\right.\\x\ge4\end{matrix}\right.\Leftrightarrow x\ge6}\)
\(x^2-4x-12\le x^2-8x+16\\ \Leftrightarrow4x\le28\Leftrightarrow x\le7\)
Vậy \(6\le x\le7\)
b) \(\left(x-2\right)\sqrt{x^2+4}\le x^2-4\Leftrightarrow\left(x-2\right)\left(x+2-\sqrt{x^2+4}\right)\ge0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2\\x+2\ge\sqrt{x^2+4}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2\\x+2\le\sqrt{x^2+4}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ge2\\x\le0\end{matrix}\right.\)
