\(\left(2x+1\right)\left(3x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy...
ta có:
(2x+1)(3x-2)=0
\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy...
\(\left(2x+1\right)\left(3x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\3x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=-1\\3x=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{3}{2}\end{matrix}\right.\)
\(\left(2x+1\right)\left(3x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\3x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=-1\\3x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\)
Ta có :
(2x + 1) (3x - 2) =0
\(\Rightarrow\left\{{}\begin{matrix}2x+1=0\\3x-2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{2}{3}\end{matrix}\right.\)
Vậy x = \(-\dfrac{1}{2}\) và y = \(\dfrac{2}{3}\)
