Question

5^x-1 ≥ 5^{x^2 - x- 9}

Answer 1

Ta có:\(5^{x-1}\ge5^{x^2-x-9}\)

   \(\Leftrightarrow x-1\ge x^2-x-9\)

   \(\Leftrightarrow x^2-2x-8\le0\)

   \(\Leftrightarrow\left(x-4\right)\left(x+2\right)\le0\)

   \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-4\le0\\x+2\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-4\ge0\\x+2\le0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le4\\x\ge-2\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge4\\x\le-2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow-2\le x\le4\)