a, \(\dfrac{15}{n-1}\); n∈Z
\(\dfrac{15\left(n-1\right)}{n-1}=\dfrac{15n-15}{n-1}\)
=> Ư(15)={\(\pm1;\pm3;\pm5;\pm15\)}
| n-1 | -15 | -5 | -3 | -1 | 1 | 3 | 5 | 15 |
| n | -14 | -4 | -2 | 0 | 2 | 4 | 6 | 16 |
| Đánh giá | t/mãn | t/mãn | t/mãn | t/mãn | t/mãn | t/mãn | t/mãn | t/mãn |
Vậy n∈{-14;-4;-2;0;2;4;6;16}
b, \(\dfrac{-21}{n+3}\) n∈Z
\(\dfrac{-21\left(n+3\right)}{n+3}=\dfrac{\left(-21n-63\right)}{n+3}\)
Ư(63)={±1;±3;±7;±9;±21;±63}
| n+3 | -63 | -21 | -9 | -7 | -3 | -1 | 1 | 3 | 7 | 9 | 21 | 63 |
| n | -66 | -24 | -12 | -10 | -6 | -4 | -2 | 0 | 4 | 6 | 18 | 60 |
| Đ/gia | t/mãn | t/mãn | t/mãn | t/mãn | t/mãn | t/mãn | t/mãn | t/mãn | t/mãn | t/mãn | t/mãn | t/mãn |
Vậy n∈{-66;-24;-12;-10;-6;-4;-2;0;4;6;18;60}
\(\dfrac{2n+7}{n-2};n\inℤ\\ \Rightarrow\dfrac{\left(2n-4\right)+7+2}{n-2}=\dfrac{2\left(n-2\right)+9}{n-2}=2+\dfrac{9}{n-2}\)
\(\LeftrightarrowƯ\left(9\right)=\left\{\pm1;\pm3;\pm9\right\}\)
Ta có bảng sau:
| n-2 | -9 | -3 | -1 | 1 | 3 | 9 |
| n | -7 | -1 | 1 | 3 | 5 | 11 |
| Đ/gia | t/mãn | t/mãn | t/mãn | t/mãn | t/mãn | t/mãn |
Vậy n={-7;-1;1;3;5;11}
