\(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy...
(x+\(\dfrac{1}{2}\) ).(\(\dfrac{2}{3}\) -2x)=0\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\2x=\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\) \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vì (x+\(\dfrac{1}{2}\)).(\(\dfrac{2}{3}\)-2.x)=0
nên x+\(\dfrac{1}{2}\)=0 hoặc \(\dfrac{2}{3}\)-2x=0
Với x+\(\dfrac{1}{2}\)=0
x=0-\(\dfrac{1}{2}\)
x=\(\dfrac{-1}{2}\)
Với \(\dfrac{2}{3}\)-2x=0
2x=\(\dfrac{2}{3}\)-0
2x=\(\dfrac{2}{3}\)
x=\(\dfrac{2}{3}\):2
x=\(\dfrac{1}{3}\)
Vậy x=\(\dfrac{-1}{2}\)và \(\dfrac{1}{3}\)
\(\dfrac{1}{2}\)