Question

Cho n \(\varepsilon\)\(ℕ^∗\). Tìm n  đê A \(\varepsilon\)\(ℤ\)

A = \(\frac{n+8}{2n-5}\)

Answer 1

n = { 3, -3 , -8

Answer 2

Để \(A\in Z\Leftrightarrow\left(n+8\right)⋮\left(2n-5\right)\)

Giả sử\(\left(n+8\right)⋮\left(2n-5\right)\)

\(\Leftrightarrow2\left(n+8\right)⋮\left(2n-5\right)\)

\(\Leftrightarrow2n+16⋮\left(2n-5\right)\)

\(\Leftrightarrow2n-5+21⋮\left(2n-5\right)\)

Do \(2n-5⋮2n-5\)

\(\Rightarrow21⋮\left(2n-5\right)\)

\(\Rightarrow\left(2n-5\right)\inƯ\left(21\right)\)

Ta có bảng sau:

2n-5	-21	-7	-3	-1	1	3	7	21
2n	-16	-2	2	4	6	8	12	26
n	-8	-1	1	2	3	4	6	13

Do \(n\inℕ^∗\Rightarrow n\in\left\{1;2;3;4;6;13\right\}\)