Question

Cho A = \(\dfrac{2n + 1}{n + 3} - \dfrac{n - 5}{n + 3}. \)

Tìm các số nguyên n để A nhận giá trị nguyên.

Answer 1

ta có:

A = \(\frac{2n+1}{n+3}-\frac{n+5}{n+3}\)

=> A = \(\frac{2n+6-5}{n+3}-\frac{n+3-8}{n+3}\)

=> A = \(\frac{2\left(n+3\right)-5}{n+3}-\left(1-\frac{8}{n+3}\right)\)

=> A = \(2-\frac{5}{n+3}-1+\frac{8}{n+3}\)

=> A = \(2-1-\frac{5}{n+3}+\frac{8}{n+3}\)

=> A = \(1-\frac{13}{n+3}\)

để A nguyên => \(\frac{13}{n+3}\) nguyên

=> \(13⋮n+3\)

=> \(n+3\inƯ_{\left(13\right)}=\left\{1;-1;13;-13\right\}\)

ta có bảng sau:

n+3	1	-1	13	-13
n	-4	-5	10	-16

vậy n = {-4; -5; 10; -16}