a) \(\frac{1}{2}x+\frac{2}{3}x-1=-3\frac{1}{3}\)
\(\Leftrightarrow\frac{7}{6}x=-\frac{7}{3}\)
\(\Leftrightarrow x=-2\)
Vậy x = -2
b) \(\left(x-3\right)\left(4-5x\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\4-5x=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{4}{5}\\x=-3\end{matrix}\right.\)
Vậy \(x\in\left\{3;\frac{4}{5};-3\right\}\)