Answer

`|x - 1/2| - \sqrt{1/9} = \sqrt{1/4}`

`=> |x - 1/2| - \sqrt{(1/3)^{2}} = \sqrt{(1/2)^{2}}`

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

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

`=> |x - 1/2| = 1/6`

`=>` $\left[\begin{matrix} x - \dfrac{1}{2} = \dfrac{1}{6}\\ x - \dfrac{1}{2} = \dfrac{-1}{6}\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = \dfrac{1}{6} + \dfrac{1}{2}\\ x = \dfrac{-1}{6} + \dfrac{1}{2}\end{matrix}\right.$

`=>` $\left[\begin{matrix} x = \dfrac{2}{3}\\ x = \dfrac{1}{3}\end{matrix}\right.$

$\text{Vậy}$ `x \in {2/3 ; 1/3}`

Mình ko chép lại đề nha!

l x - $\frac{1}{2}$ l -  $\frac{1}{3}$ =  $\frac{1}{2}$ 

l x - $\frac{1}{2}$ l                          =  $\frac{1}{2}$ +  $\frac{1}{3}$ 

l x - $\frac{1}{2}$ l                          = $\frac{5}{6}$ 

TH1: x - $\frac{1}{2}$ = $\frac{5}{6}$ 

x = $\frac{5}{6}$  + $\frac{1}{2}$ 

x = $\frac{4}{3}$

TH2: x - $\frac{1}{2}$ = $\frac{-5}{6}$ 

x = $\frac{-5}{6}$ + $\frac{1}{2}$

x = $\frac{-1}{3}$ 

Vậy x ∈ { $\frac{4}{3}$ ; $\frac{-1}{3}$