Question

Giải các bất phương trình sau:

a) \(\sqrt{x^2-3x-10}>x-2\)

Answer 1

\(\sqrt{2x^2-3x-10}< x-2\Leftrightarrow\left\{{}\begin{matrix}x-2>0\\x^2-3x-10\ge0\\x^2-3x-10< \left(x-2\right)^2\end{matrix}\right.\left\{{}\begin{matrix}x>2\\\left[{}\begin{matrix}x\ge5\\x\le-2\end{matrix}\right.\\x^2-3x-10< \left(x-2\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge5\\x^2-3x-10< x^2-4x+4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge5\\x< 14\end{matrix}\right.\Leftrightarrow5\le x< 14\Rightarrow x\in\text{[ 5 ; 14 )}\)