2: \(\left(4x+1\right)\left(-2x+\dfrac{1}{3}\right)=0\)
=>\(\left[{}\begin{matrix}4x+1=0\\-2x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-1\\2x=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=\dfrac{1}{6}\end{matrix}\right.\)
5: \(\left(3-2x\right)\left(\dfrac{4}{7}x+2\right)=0\)
=>\(\left[{}\begin{matrix}3-2x=0\\\dfrac{4}{7}x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\\dfrac{4}{7}x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2:\dfrac{4}{7}=-\dfrac{14}{4}=-\dfrac{7}{4}\end{matrix}\right.\)
8: \(\left(\dfrac{1}{3}x-\dfrac{7}{9}\right)\left(\dfrac{6}{5}-\dfrac{3}{2}:x\right)=0\)
=>\(\left[{}\begin{matrix}\dfrac{1}{3}x-\dfrac{7}{9}=0\\\dfrac{6}{5}-\dfrac{3}{2}:x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{3}x=\dfrac{7}{9}\\\dfrac{3}{2}:x=\dfrac{6}{5}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{7}{9}:\dfrac{1}{3}=\dfrac{7}{3}\\x=\dfrac{3}{2}:\dfrac{6}{5}=\dfrac{3}{2}\cdot\dfrac{5}{6}=\dfrac{15}{12}=\dfrac{5}{4}\end{matrix}\right.\)
