13, `(2-x)/3` = `(4+2x)/5`
`=>` 5 ( 2 - x ) = 3 ( 4 + 2x )
`=>` 10 - 5x = 12 + 6x
`=>` 10 - 12 = 6x + 5x
`=>` -2 = 11x
`=>` x = `(-2)/11`
Vậy x = `(-2)/11`
15, | x - 1 | + | 3 - 2x | = 0
Ta có : | x - 1 | ≥ 0 ∀ x
| 3 - 2x | ≥ 0 ∀ x
`=>` | x - 1 | + | 3 - 2x | ≥ 0 ∀ x
Để | x - 1 | + | 3 - 2x | = 0 `=>` $\left \{ {{x-1=0 ( 1 ) } \atop {3-2x=0(2)}} \right.$
( 1 ) `=>` x = 1
( 2 ) `=>` x = `3/2`
Vậy x ∈ { 1 ; `3/2` }
