Question

[Huyền Nguyễn]

Bài 1: Tìm nϵZ để:

a)(2n+3)⋮(n-1)                          c)(2n²+n+3)⋮(2n+1)
b)(n²-2n+4)⋮(n+1)                    d)(2n²-n+2)⋮(n+2)

Bài 2:

1)Chứng tỏ rằng: A=\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{4041}{2020^2.2021^2}+\dfrac{4043}{2021^2.2022^2}< 1\) Rồi suy ra A không phải số nguyên

2)Cho 200 tia chung gốc không có tia nào trùng nhau hỏi tạo thành bao nhiêu góc.

 

Answer 1

[Nguyễn Lê Phước Thịnh]

bài 1:

a: \(2n+3⋮n-1\)

=>\(2n-2+5⋮n-1\)

=>\(5⋮n-1\)

=>\(n-1\in\left\{1;-1;5;-5\right\}\)

=>\(n\in\left\{2;0;6;-4\right\}\)

b: \(n^2-2n+4⋮n+1\)

=>\(n^2+n-3n-3+7⋮n+1\)

=>\(7⋮n+1\)

=>\(n+1\in\left\{1;-1;7;-7\right\}\)

=>\(n\in\left\{0;-2;6;-8\right\}\)

c: \(2n^2+n+3⋮2n+1\)

=>\(n\left(2n+1\right)+3⋮2n+1\)

=>\(3⋮2n+1\)

=>\(2n+1\in\left\{1;-1;3;-3\right\}\)

=>\(n\in\left\{0;-1;1;-2\right\}\)
d: \(2n^2-n+2⋮n+2\)

=>\(2n^2+4n-5n-10+12⋮n+2\)

=>\(12⋮n+2\)

=>\(n+2\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)

=>\(n\in\left\{-1;-3;0;-4;1;-5;2;-6;4;-8;10;-14\right\}\)

Answer 2

[Huyền Nguyễn]

a)(n²-3n+1)⋮(n-2)
Vì (n-2)⋮(n-2)
⇒n.(n-2)⋮(n-2)
⇒[(n²-3n+1)-n.(n-2)]⋮(n-2)
⇒[(n²-3n+1)-(n²-2n)]⋮(n-2)
⇒[n²-3n+1-n²+2n⁰ ]⋮(n-2)
⇒(-n+1):(n-2)
⇒-(n-1)⋮(n-2)
⇒(n-2+1)⋮(n-2)
 Vì (n-2)⋮(n-2)
⇒1⋮(n-2)
 Vì n nguyên
⇒(n-2)ϵƯ(1)={-1;1}Ư