d, \(\left(x^{2+4x}\right)\left(5-x\right)< 0\Leftrightarrow\left[{}\begin{matrix}x^{2+4x}< 0\\5-x< 0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2.4x< 0\\x>5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< 0\\x>5\end{matrix}\right.\)

Vì \(x^2\ge0\forall x\in Q\Rightarrow x^2.4x< 0\Leftrightarrow4x< 0\Leftrightarrow x< 0\)

e, \(\dfrac{x}{x+1}>0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x+1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x>1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x< -1\end{matrix}\right.\end{matrix}\right.\)

Vậy = khi x > 1 hoặc x < -1

f, \(\dfrac{2x-1}{2-x}\le0\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\\left\{{}\begin{matrix}2x-1>0\\2-x>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-1< 0\\2-x< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\\left\{{}\begin{matrix}x>\dfrac{1}{2}\\x< 2\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{1}{2}\\x>2\end{matrix}\right.\end{matrix}\right.\)

Vậy --... khi x = 1/2(thỏa mãn) hoặc 1/2