1.
a) Tìm số dư trong phép chia f(x) = ( $x^{2}$ + 8x+ 15)(x+2020)+2021 cho x + 5
b) Giải phương trình:
$\frac{390-x}{2016}$ + $\frac{388-x}{2018}$ +$\frac{386-x}{2020}$ + $\frac{384-x}{2020}$ = -4
2. Chứng minh rằng với mọi số tự nhiên lẻ n ta luôn có n ³ – n chia hết cho 24
$1.$
$a,$
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 …………………