`(2/3 -x)^2 - 1/9=0`
`(2/3-x)^2 = 0+1/9`
`(2/3-x)^2 = 1/9`
`(2/3-x)^2 = (1/3)^2`
`@TH1`
`2/3-x = 1/3`
`x=2/3 -1/3`
`x=1/3`
`@TH2:`
`2/3-x=-1/3`
`x=2/3-(-1/3)`
`x=2/3+1/3`
`x=3/3`
`x=1`
Vậy`x = {1/3;1}`
\(\rightarrow\left(\dfrac{2}{3}-x\right)^2=\dfrac{1}{9}\\ \left(\dfrac{2}{3}-x\right)^2=\left(\dfrac{1}{3}\right)^2\\ \rightarrow\dfrac{2}{3}-x=\dfrac{1}{3}\\ x=\dfrac{2}{3}-\dfrac{1}{3}\\ x=\dfrac{1}{3}\)