Để \(x\inℤ\)khi \(\frac{3x+12}{2x+4}\)là số nguyên
hay \(3x+12⋮2x+4\Leftrightarrow6x+24⋮2x+4\)
\(\Leftrightarrow3\left(2x+4\right)+12⋮2x+4\Rightarrow2n+4\inƯ\left(12\right)=\left\{\pm1;\pm2;\pm3;\pm4;\pm6;\pm12\right\}\)
2n + 4 | 1 | -1 | 2 | -2 | 3 | -3 | 4 | -4 | 6 | -6 | 12 | -12 |
n | -3/2 ( ktm ) | -5/2 ( ktm ) | -1 | -3 | -1/2 ( ktm ) | -7/2 ( ktm ) | 0 | -4 | 1 | -5 | 4 | -8 |