\(A=\frac{3}{n+2}\)
a) \(\Leftrightarrow n+2\inƯ\left(3\right)\)
\(\Leftrightarrow n+2\in\left\{\pm1;\pm3\right\}\)
+) \(n+2=1\Leftrightarrow n=-1\)
+) \(n+2=-1\Leftrightarrow n=-3\)
+) \(n+2=3\Leftrightarrow n=1\)
+) \(n+2=-3\Leftrightarrow n=-5\)
b) \(A=\frac{3}{2};A=\frac{3}{2+2}=\frac{3}{4};A=\frac{3}{-7+2}=\frac{3}{-5}\)
\(A=\frac{3}{n+2}\)
Để A là phân số => \(n+2\ne0\)=> \(n\ne-2\)
n = 0 => \(A=\frac{3}{0+2}=\frac{3}{2}\)
n = 2 => \(A=\frac{3}{2+2}=\frac{3}{4}\)
n = -7 => \(A=\frac{3}{-7+2}=\frac{3}{-5}=\frac{-3}{5}\)
\(A=\frac{3}{n+2}\)
\(\Rightarrow n+2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
Lập bảng
n + 2 | 1 | -1 | 3 | -3 |
n | -1 | -3 | 1 | -5 |