Question

Answer 1

a, Để A là số hữu tỉ khi n khác -3 

b, \(A=\dfrac{n+3-2}{n+3}=1-\dfrac{2}{n+3}\Rightarrow n+3\inƯ\left(-2\right)=\left\{\pm1;\pm2\right\}\)

n+3	1	-1	2	-2
n	-2	-4	-1	-5

 

c, \(A=1-\dfrac{2}{n+3}=\dfrac{5}{6}\Leftrightarrow\dfrac{n+1}{n+3}=\dfrac{5}{6}\Rightarrow6n+6=5n+15\Leftrightarrow n=9\)