$1.$

$a,$

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 $n^3-n⋮24(đpcm)$

Câu 2: 

n^3-3n^2-n+3
=n^2(n-3)-(n-3)
=(n-3)(n^2-1)

= (n-3)(n-1)(n+1)

Với n lẻ =>(n-3)(n-1)(n+1) là tích ba số chẵn liên tiếp

=>(n-3)(n-1)(n+1):24

Câu 1:  Xét (x2+8x+15)(x+2020)=(x+5)(x+3)(x+2020)⋮x+5(x2+8x+15)(x+2020)=(x+5)(x+3)(x+2020)⋮x+5

⇒(x+5)(x+3)(x+2020)+2021:x+5⇒(x+5)(x+3)(x+2020)+2021:x+5dư 20212021

Hay (x):x+5f(x):x+5 dư 20212021

Vậy .....................

b,b,

390−x2016+388−x2018+386−x2020+384−x2022=−4390−x2016+388−x2018+386−x2020+384−x2022=−4

⇔390−x2016+1+388−x2018+1+386−x2020+1+384−x2022+1=0⇔390−x2016+1+388−x2018+1+386−x2020+1+384−x2022+1=0

⇔390−x+20162016+388−x+20182018+386−x+20202020+384−x+20222022=0⇔390−x+20162016+388−x+20182018+386−x+20202020+384−x+20222022=0

⇔2406−x2016+2406−x2018+2406−x2020+2406−x2022=0⇔2406−x2016+2406−x2018+2406−x2020+2406−x2022=0

⇔(2406−x)(12016+12018+12020+12022)=0⇔(2406−x)(12016+12018+12020+12022)=0

⇔2406−x=0(⇔2406−x=0(vì 12016+12018+12020+12022≠0)12016+12018+12020+12022≠0)

⇔x=2406⇔x=2406

Vậy .....................