`a) | x - 1| = 1 - x`
Vì : $|x-1| ≥ 0 ∀ x$
$⇒ 1-x ≥ 0 ⇔ x ≤ 1$
Vậy $x ≤ 1$
$b$ ) `| 2x + 4,5| - |x - 2,7| = 0`
`⇒` `|2x+4,5| = |x-2,7|`
$⇒$ \(\left[ \begin{array}{l}2x+4,5 = x- 2,7\\2x+4,5 = -x + 2,7\end{array} \right.\)
$⇒$ \(\left[ \begin{array}{l}x=-7,2\\x=-0,6\end{array} \right.\)
Vậy `x` `∈` `{-7,2;-0,6}`
$c$) `| x + 13/14 | + | x - 3/7 | = 0`
Vì : ` |x + 13/14 | ; | x - 3/7 | ≥ 0 ∀ x`
`⇒` $\left\{\begin{matrix}x = \dfrac{-13}{14} & \\x = \dfrac{3}{7} & \end{matrix}\right.$ ($KTM$)
Vậy $x$ $∈$ $∅$
