\(\left(2x-1\right)\left(x+\dfrac{2}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-2}{3}\end{matrix}\right.\)
Vậy .....
\(\left(2x-1\right)\left(x+\dfrac{2}{3}\right)=0\)
Vậy \(x-1=\dfrac{2}{3}\)
\(x=\dfrac{2}{3}+1\)
\(x=\dfrac{5}{3}\)
\(\left(2x-1\right)\left(x+\dfrac{2}{3}\right)=0 \)
=> \(\left[{}\begin{matrix}2x-1=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\)
Nếu \(2x-1=0 \)
\(\Rightarrow2x=0+1\)
\(2x=1\)
\(x=1:2\)
\(x=\dfrac{1}{2}\)
Nếu \(x+\dfrac{2}{3}=0\)
\(x=0-\dfrac{2}{3}\)
\(x=\dfrac{-2}{3}\)
Vậy \(x\in\left\{\dfrac{1}{2};\dfrac{-2}{3}\right\}\)
\(\left(2x-1\right)\left(x+\dfrac{2}{3}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{2}{3}\end{matrix}\right.\)
