a, Với x = 1 ta có :
\(A=\frac{3.1+2}{1-3}=\frac{3+2}{1-3}=\frac{5}{-2}=-\frac{5}{2}\)
Với x = 2 ta có
\(A=\frac{3.2+2}{2-3}=\frac{6+2}{-1}=-8\)
Với x = \(\frac{5}{2}\)
\(A=\frac{3.\frac{5}{2}+2}{\frac{5}{2}-3}=\frac{\frac{15}{2}+2}{-\frac{1}{2}}=-\frac{\frac{19}{2}}{\frac{1}{2}}=-19\)
b, \(A=\frac{3x+2}{x-3}=3+\frac{11}{x-3}\) ĐK \(x\ne3\)
Để A nguyên \(\Rightarrow x-3\inƯ\left(11\right)=\left(-11;-1;1;11\right)\)
\(x-3=-11\Rightarrow x=-8\)
\(x-3=-1\Rightarrow x=2\)
\(x-3=1\Rightarrow x=4\)
\(x-3=11\Rightarrow x=14\)
Vậy \(x=\left(-8;2;4;14\right)\)thì A nguyên