Question

giải baats phương trình

\(\left(n^2-3n\right)\sqrt{2n^2-3n-2}>=0\)

Answer 1

\(\left\{{}\begin{matrix}2n^2-3n-2\ge0\\n^2-3n\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(n-2\right)\left(2n+1\right)\ge0\\n\left(n-3\right)\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}n\le-\dfrac{1}{2}\\n\ge2\end{matrix}\right.\\\left[{}\begin{matrix}n\le0\\n\ge3\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}n\le-\dfrac{1}{2}\\n\ge3\end{matrix}\right.\)