\(\left(3x-1\right)\left(\frac{2}{3}x+\frac{1}{5}\right)\le0\)
\(\Rightarrow\left[\begin{array}{nghiempt}3x-1\le0\\\frac{2}{3}x+\frac{1}{5}\le0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}3x\le1\\\frac{2}{3}x\le-\frac{1}{5}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x\le\frac{1}{3}\\x\le-\frac{3}{10}\end{array}\right.\)
\(\Rightarrow x\le\frac{1}{3}\left(tm\right)\)
Vậy để \(\left(3x-1\right)\left(\frac{2}{3}x+\frac{1}{5}\right)\le0\) thì \(x\le\frac{1}{3}\)