`|5/2 - x| = 13/10`
`⇒` \(\left[ \begin{array}{l}\dfrac{5}{2}-x=\dfrac{13}{10}\\\dfrac{5}{2}-x=-\dfrac{13}{10}\end{array} \right.\)
$⇒$ \(\left[ \begin{array}{l}x= \dfrac{6}{5}\\x = \dfrac{19}{5}\end{array} \right.\)
Vậy `x` `∈` `{19/5;6/5}`
`|x+3/4| + 1/3 = 0`
`⇔ |x+3/4| = -1/3`
Vì : `|x+3/4| ≥ 0 ∀ x`
`⇒ x ∈`∅
Vậy `x` `∈` ∅
`|x-3/2| + |5/2 - x| = 0`
Vì : `|x-3/2|;|5/2-x| ≥ 0 ∀ x`
`⇒` `x-3/2 = 5/2 - x = 0`
`⇒` `x=3/2;x = 5/2` (Không thỏa mãn)
Vậy `x` `∈` $\varnothing$.
