\(\left(x-1\right)\left(3x+2\right)\left(2y-3\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-1=0\\3x+2=0\\\left(2y-3\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=1\\3x=-2\\2y-3=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=1\\x=\dfrac{-2}{3}\\2y=3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=1\\x=\dfrac{-2}{3}\\y=\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x\in\left\{1;\dfrac{-2}{3}\right\}\\y\in\left\{\dfrac{3}{2}\right\}\end{matrix}\right.\)
\(\left(x-1\right)\left(3x+2\right)\left(2y-3\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3x+2=0\\\left(2y-3\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{2}{3}\\y=\dfrac{3}{2}\end{matrix}\right.\)
Vậy ...
