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 \(\dfrac{1}{3}\)
\(\dfrac{1}{2}\)
