\(\left(2x-\dfrac{2}{3}\right)^2-1=\dfrac{7}{9}\)
\(\Leftrightarrow\left(2x-\dfrac{2}{3}\right)^2=\dfrac{16}{9}=\left(\text{±}\dfrac{4}{3}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x=2\\2x-\dfrac{2}{3}=-\dfrac{4}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=2\\2x=-\dfrac{2}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)
`( 2x - 2/3 )^2 - 1 = 7/9`
`=> ( 2x - 2/3 )^2 = 7/9+1`
`=> ( 2x - 2/3 )^2 = 16/9 = ( 4/3 )^2 = ( -4/3 )^2`
`=> 2x - 2/3 = 4/3` hoặc `2x - 2/3 = -4/3`
`=> 2x = 6/3 = 2` hoặc `2x = -2/3`
`=> x = 1` hoặc `x = -1/3`
Vậy `x in { 1;-1/3}`
`(2x-2/3)^2-1=7/9`
`(2x-2/3)^2=7/9+1`
`(2x-2/3)^2=16/9`
`(2x-2/3)^2=(4/3)^2` hoặc `(2x-2/3)^2=(-4/3)^2`
`@TH1:2x-2/3=4/3`
`=>2x=4/3+2/3=2`
`=>x=2:2=1`
`@TH2:2x-2/3=-4/3`
`=>2x=-4/3+2/3=-2/3`
`=>x=-2/3:2=-1/3`
Vậy `x=1` hoặc `x=-1/3`