Chia đa thức \(\left(n^2-2\right):\left(n-3\right)=\left(n+3\right)\)dư 7
\(\Rightarrow A=Q+\frac{R}{B}=n+3+\frac{7}{n-3}\)
\(\Rightarrow\left(n-3\right)\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
Ta có bảng sau:
n-3 | 1 | -1 | 7 | -7 |
n | 4 (t/m) | 2 (t/m) | 10 (t/m) | -4 (t/m) |