Để A là số nguyên thì 3n+9⋮n-4
=>3n-12+21⋮n-4
=>21⋮n-4
=>n-4∈{1;-1;3;-3;7;-7;21;-21}
=>n∈{5;3;7;1;11;-3;25;-17}
Khi n=5 thì \(A=\frac{3\cdot5+9}{5-4}=\frac{15+9}{1}=24\)
Khi n=3 thì \(A=\frac{3\cdot3+9}{3-4}=\frac{9+9}{-1}=-18\)
Khi n=7 thì \(A=\frac{3\cdot7+9}{7-4}=\frac{21+9}{3}=\frac{30}{3}=10\)
Khi n=1 thì \(A=\frac{3\cdot1+9}{1-4}=\frac{12}{-3}=-4\)
Khi n=11 thì \(A=\frac{3\cdot11+9}{11-4}=\frac{33+9}{7}=\frac{42}{7}=6\)
Khi n=-3 thì \(A=\frac{3\cdot\left(-3\right)+9}{-3-4}=0\)
Khi n=25 thì \(A=\frac{3\cdot25+9}{25-4}=\frac{75+9}{21}=\frac{84}{21}=4\)
Khi n=-17 thì \(A=\frac{3\cdot\left(-17\right)+9}{-17-4}=\frac{-51+9}{-21}=\frac{-42}{-21}=2\)
Để B nguyên thì 6n+5⋮2n-1
=>6n-3+8⋮2n-1
=>8⋮2n-1
=>2n-1∈{1;-1}
=>2n∈{2;0}
=>n∈{1;0}
Khi n=1 thì \(B=\frac{6\cdot1+5}{2\cdot1-1}=\frac{11}{1}=11\)
Khi n=0 thì \(B=\frac{6\cdot0+5}{2\cdot0-1}=\frac{5}{-1}=-5\)