a) \(x^2-5x\ge0\Leftrightarrow x\left(x-5\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge0\\x-5\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x\le0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge5\\x\le0\end{matrix}\right.\)
b) Thiếu vế phải rồi ạ
c) Do \(x^2+1\ge1>0\forall x\)
\(\Rightarrow-x^2+3x+4\le0\Leftrightarrow-\left(x^2-3x+\dfrac{9}{4}\right)+\dfrac{25}{4}\le0\)
\(\Leftrightarrow-\left(x-\dfrac{3}{2}\right)^2+\dfrac{5}{4}\le0\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2\ge\dfrac{5}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}\ge\dfrac{\sqrt{5}}{2}\\x-\dfrac{3}{2}\le-\dfrac{\sqrt{5}}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{3+\sqrt{5}}{2}\\x\le\dfrac{3-\sqrt{5}}{2}\end{matrix}\right.\)
