$\\$
`a,`
`A=(3x+3)/(x^3+x^2+x+1)`
`->A=(3(x+1))/((x^3+x^2)+(x+1))`
`->A=(3(x+1))/(x^2(x+1)+(x+1))`
`->A=(3(x+1))/((x+1)(x^2+1))`
`->A=3/(x^2+1)`
Để `A` nguyên
`->3` chia hết cho `x^2+1`
`->x^2+1 ∈ Ư (3)={1;-1;3;-3}`
$\bullet$ `x^2+1=1 ->x^2=0 ->x=0` (tm)
$\bullet$ `x^2+1=-1 ->x^2=-2 ->x=∅`
$\bullet$ `x^2+1=3 ->x^2=2 ->x=±\sqrt{2}` (tm)
$\bullet$ `x^2+1=-3 ->x^2=-4->x=∅`
Vậy `x ∈ {0; ±\sqrt{2}}` để `A` nguyên
$\\$
`b,`
`A=3/(x^2+1)`
Vì `x^2≥0∀x`
`->x^2+1 ≥1∀x`
`->3/(x^2+1) ≤3∀x`
`->A≤3∀x`
Dấu "`=`" xảy ra khi :
`x^2=0↔x=0`
Vậy `max A=3 ↔x=0`
