Bài 14:
a: Ta có: \(A=\left(2x-5\right)\left(3x-\dfrac{5}{3}\right)\)
Để A<0 thì \(\left\{{}\begin{matrix}3x-\dfrac{5}{3}>0\\2x-5< 0\end{matrix}\right.\Leftrightarrow\dfrac{5}{9}< x< \dfrac{5}{2}\)
b: Để \(x^2+2020x< 0\) thì \(x\left(x+2020\right)< 0\)
hay \(-2020< x< 0\)