$1.$
$a,$
Xét $(x^2+8x+15)(x+2020)=(x+5)(x+3)(x+2020)⋮x+5$
$⇒(x+5)(x+3)(x+2020)+2021:x+5$ dư $2021$
Hay $f(x):x+5$ dư $2021$
Vậy .....................
$b,$
$\frac{390-x}{2016}+\frac{388-x}{2018}+\frac{386-x}{2020}+\frac{384-x}{2022}=-4$
$⇔\frac{390-x}{2016}+1+\frac{388-x}{2018}+1+\frac{386-x}{2020}+1+\frac{384-x}{2022}+1=0$
$⇔\frac{390-x+2016}{2016}+\frac{388-x+2018}{2018}+\frac{386-x+2020}{2020}+\frac{384-x+2022}{2022}=0$
$⇔\frac{2406-x}{2016}+\frac{2406-x}{2018}+\frac{2406-x}{2020}+\frac{2406-x}{2022}=0$
$⇔(2406-x)(\frac{1}{2016}+\frac{1}{2018}+\frac{1}{2020}+\frac{1}{2022})=0$
$⇔2406-x=0($vì $\frac{1}{2016}+\frac{1}{2018}+\frac{1}{2020}+\frac{1}{2022}≠0)$
$⇔x=2406$
Vậy .....................
$2.$
Xét $n^3-n=n(n^2-1)=n(n-1)(n+1)$
Do $n;n-1;n+1$ là 3 số tự nhiên liên tiếp $⇒n(n-1)(n+1)⋮3$ $(1)$
Mà $n$ là số lẻ $⇒n-1;n+1$ là các số chẵn liên tiếp
$⇒$ 1 trong 2 số chia hết cho 4
$⇒(n-1)(n+1)⋮2.4=8$ $(2)$
Từ $(1)$ và $(2)$ $⇒n(n-1)(n+1)⋮3.8=24($vì $(3;8)=1)$
Hay $n^3-n⋮24(đpcm)$
