a, ĐKXĐ:\(n-3\ne0\Leftrightarrow n\ne3\)
b, Thay n=-3 vào B ta có:
\(B=\dfrac{14}{n-3}=\dfrac{14}{-3-3}=\dfrac{14}{-6}=\dfrac{-7}{3}\)
c, Để \(B=\dfrac{13}{n}\left(n\ne0\right)\)
\(\Leftrightarrow\dfrac{14}{n-3}=\dfrac{13}{n}\\ \Leftrightarrow14n=13n-39\\ \Leftrightarrow n=-39\left(tm\right)\)
d, \(\dfrac{14}{n-3}\in Z\)
\(\Leftrightarrow14⋮\left(n-3\right)\\ \Leftrightarrow n-3\inƯ\left(14\right)=\left\{-14;-7;-2;-1;1;2;7;14\right\}\)
Ta có bảng:
| n-3 | -14 | -7 | -2 | -1 | 1 | 2 | 7 | 14 |
| n | -11 | -4 | 1 | 2 | 4 | 5 | 10 | 17 |
Vậy \(n\in\left\{-11;-4;1;2;4;5;10;17\right\}\)
Đúng 1
Bình luận (0)
