Đặt \(D=\frac{2a+9}{a+3}+\frac{5a+17}{a+3}-\frac{3a}{a+3}\)
\(=\frac{2a+9+5a+17-3a}{a+3}\)
\(=\frac{4a+26}{a+3}=\frac{4\left(a+3\right)+14}{a+3}=4+\frac{14}{a+3}\)
\(\Rightarrow14⋮a+3\)
\(\Rightarrow a+3\inƯ\left(14\right)\)
Đến đây làm nốt
Đặt \(A=\frac{2a+9}{a+3}+\frac{5a+17}{a+3}-\frac{3a}{a+3}\)
\(\Rightarrow A=\frac{\left(2a+9\right)+\left(5a+17\right)-3a}{a+3}=\frac{4a+26}{a+3}=\frac{4a+12+14}{a+3}\)
\(=\frac{4\left(a+3\right)+14}{a+3}=4+\frac{14}{a+3}\)
Vì \(4\inℤ\)\(\Rightarrow\)Để A nguyên thì \(14⋮\left(a+3\right)\)\(\Rightarrow a+3\inƯ\left(14\right)=\left\{\pm1;\pm2;\pm7;\pm14\right\}\)
\(\Rightarrow a\in\left\{-17;-10;-5;-4;-2;-1;4;11\right\}\)
Vậy \(a\in\left\{-17;-10;-5;-4;-2;-1;4;11\right\}\)
