Question

Giải bất phương trình: \(\sqrt{x^2-4x}>x-3\)

Answer 1

\(\sqrt{x^2-4x}>x-3\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3< 0\\x^2-4x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3\ge0\\x^2-4x>\left(x-3\right)^2\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 3\\\left[{}\begin{matrix}x\le0\\x\ge4\end{matrix}\right.\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge3\\x>\dfrac{9}{2}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\le0\\x>\dfrac{9}{2}\end{matrix}\right.\)