A=\(\dfrac{4n^2-19}{2n^2+3}=\dfrac{4n^2+6-25}{2n^2+3}=\dfrac{\left(4n^2+6\right)-25}{2n^2+3}=\dfrac{2\left(2n^2+3\right)-25}{2n^2+3}=\dfrac{2\left(2n^2+3\right)}{2n^2+3}-\dfrac{25}{2n^2+3}=2-\dfrac{25}{2n^2+3}\)Để A nguyên \(\Rightarrow2-\dfrac{25}{2n^2+3}\) nguyên mà 2 nguyên \(\Rightarrow\)\(\dfrac{25}{2n^2+3}\) \(\in\)Z\(\Rightarrow25⋮2n^2+3\)
Còn lại bạn tự làm nha