`|x| + x = 1/3`

`|x| = 1/3 - x`

`⇒` \(\left[ \begin{array}{l}x=\dfrac{1}{3} - x\\x=-(\dfrac{1}{3} + x)\end{array} \right.\)

`⇒` \(\left[ \begin{array}{l}x=\dfrac{1}{3} - x\\x=-\dfrac{1}{3} + x\end{array} \right.\)

`⇒` \(\left[ \begin{array}{l}x + x=\dfrac{1}{3}\\x - x=-\dfrac{1}{3}\end{array} \right.\)

`⇒` \(\left[ \begin{array}{l}2x=\dfrac{1}{3}\\⇔x = \dfrac{1}6\\0x =-\dfrac{1}{3} (KTM)\end{array} \right.\)

Vậy `x = 1/6`

Hay nhất !

`\text{Ta có:}`

   `|x| + x = 1/3`

   `|x| = 1/3 - x`

   `x = 1/3 - x  \text{hoặc}  x = -(1/3 - x)`

   `x = 1/3 - x  \text{hoặc}  x = -1/3 + x`

   `x + x = 1/3  \text{hoặc}  x - x = -1/3`

   `2x = 1/3  \text{hoặc}  0x = -1/3`

   `2x = 1/3 : 2  \text{hoặc}  0x = -1/3`

`\text{Vậy}  x = 1/6`