`(-2/3+x)^2=1/4`

`=>(-2/3+x)^2=(1/2)^2`

Trường hợp `1:`

`-2/3+x=1/2`

`=>x=1/2-(-2/3)`

`=>x=1/2+2/3`

`=>x=3/6+4/6`

`=>x=7/6`

Trường hợp `2:`

`-2/3+x=-1/2`

`=>x=-1/2-(-2/3)`

`=>x=-1/2+2/3`

`=>x=-3/6+4/6`

`=>x=1/6`

Vậy `x in {7/6;1/6}`

Đáp án: `x = \frac{7}{6}` hoặc `x = \frac{1}{6}`

Giải thích các bước giải:

`(- \frac{2}{3} + x )^2 = \frac{1}{4}`
(Do `\frac{1}{4} = (\frac{1}{2})^2 = (\frac{-1}{2})^2`)
`<=>` \(\left[ \begin{array}{l}\Big(\dfrac{-2}{3} + x\Big)^2 = \Big(\dfrac{1}{2}\Big)^2\\\Big(\dfrac{-2}{3} + x\Big)^2 = \Big(\dfrac{-1}{2}\Big)\end{array} \right.\) 
`<=>` \(\left[ \begin{array}{l}\dfrac{-2}{3} + x = \dfrac{1}{2}\\\dfrac{-2}{3} + x = \dfrac{-1}{2}\end{array} \right.\) 
`<=>` \(\left[ \begin{array}{l}  x = \dfrac{1}{2} + \dfrac{2}{3}\\  x = \dfrac{-1}{2} + \dfrac{2}{3}\end{array} \right.\) 
`<=>` \(\left[ \begin{array}{l}  x = \dfrac{7}{6}\\  x = \dfrac{1}{6}\end{array} \right.\) 
Vậy `x = \dfrac{7}{6}` hoặc `x = \dfrac{1}{6}`