Question

Tìm n thuộc N sao cho:

a,4n+1 chia hết 2n-1

b,2n+5 chia hết n+2

c,2n+3 chia hết n-2

 

Chi tiết hộ em ạ!

Em cảm ơn rất nhiều!

Answer 1

a) \(\frac{4n+1}{2n-1}=\frac{4n-2+3}{2n-1}=\frac{2.\left(2n-1\right)+3}{2n-1}\)

\(=2+\frac{3}{2n-1}\). Vì \(2\in Z\Rightarrow\frac{3}{2n-1}\in Z\Rightarrow2n-1\inƯ\left(3\right)\)

\(\Rightarrow2n-1\in\left\{-3;-1;1;3\right\}\)

\(\Rightarrow2n\in\left\{-2;0;2;4\right\}\)

\(\Rightarrow n\in\left\{-1;0;1;2\right\}\)

b)\(\frac{2n+5}{n+2}=\frac{2n+4+1}{n+2}=\frac{2.\left(n+2\right)+1}{n+2}\)

\(=\frac{2.\left(n+2\right)}{n+2}+\frac{1}{n+2}=2+\frac{1}{n+2}\). Vì \(2\in Z\Rightarrow n+2\inƯ\left(1\right)\)

\(\Rightarrow n+2\in\left\{-1;1\right\}\)

\(\Rightarrow n\in\left\{-3;-1\right\}\)

c) \(\frac{2n-3}{n-2}=\frac{2n-4+1}{n-2}=\frac{2.\left(n-2\right)+1}{n-2}\)

\(=\frac{2.\left(n-2\right)}{n-2}+\frac{1}{n-2}=2+\frac{1}{n-2}\)

Vì \(2\in Z\Rightarrow\frac{1}{n-2}\in Z\Rightarrow n-2\inƯ\left(1\right)\)

\(\Rightarrow n-2\in\left\{-1;1\right\}\)

\(\Rightarrow n\in\left\{1;3\right\}\)

Answer 2

Ta có: \(4n+1⋮2n-1\Leftrightarrow4n-2+3⋮2n+1\)\(\Leftrightarrow2\left(2n-1\right)+3⋮2n-1\Leftrightarrow3⋮2n-1\)

\(\Rightarrow2n-1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

\(\Rightarrow2n=\left\{-2;0;2;4\right\}\)

Vì \(n\in N\)nên \(n=\left\{0;1;2\right\}\)

Answer 3

Các câu trên bỏ các giá trị của n mà có dấu "-" nha bạn. Mình chưa đọc kỹ đề

Answer 4

b) Ta có: \(2n+5⋮n+2\Leftrightarrow2n+10-5⋮n+2\)\(\Leftrightarrow2\left(n+5\right)-5⋮n+2\Leftrightarrow5⋮n+2\)

\(\Rightarrow n+2\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\)

Vì \(n\in N\)nên \(\Rightarrow n=3\)