$3$.
|$x+19$| + |$x+5$| + |$x+2011$| = $4x$
Nhận xét: |$x+19$|;|$x+5$|;|$x+2011$| ≥ $0$ ∀ $x$ ∈ $Z$
⇒|$x+19$| + |$x+5$| + |$x+2011$| ≥ $0$⇔$4x$ ≥ $0$
Mà $4$ ≥ $0$⇒$x ≥ 0$
⇒$x+19+x+5+x+2011=4x$
⇔ $3x + 2035 =4x$
⇔ $x =2035$
Vậy $x=2035$
$4$.
$a$, $n-1$ là ước của $15$
⇒$n-1$ ∈ {$±1;±3;±5;±15$}
⇔$n$ ∈ {$-14;-4;-2;0;2;4;6;16$}
Vậy $n$ ∈ {$-14;-4;-2;0;2;4;6;16$}
$b$, $2n-1$ ⋮ $n-3$
⇔$2n-1 - 2$($n-3$) ⋮ $n-3$
⇔$2n - 1 - 2n + 6$ ⋮ $n-3$
⇔ $5$ ⋮ $n-3$
⇒$n-3$ ∈ Ư($5$)={$±1;±5$}
⇔ $n$ ∈ {$-2;2;4;8$}
Vậy $n$ ∈ {$-2;2;4;8$}