\(\left(\frac{3}{4}x-\frac{9}{16}\right)\cdot\left(1,5+\frac{-3}{5}:x\right)=0\\ \Rightarrow\left[{}\begin{matrix}\frac{3}{4}x-\frac{9}{16}=0\\1,5+\frac{-3}{5}:x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{3}{4}x=\frac{9}{16}\\\frac{-3}{5}:x=-1,5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\frac{3}{4}\\x=\frac{2}{5}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{3}{4};\frac{2}{5}\right\}\)