\(...A=\dfrac{x+19+2x-6}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x+13}{\left(x-3\right)\left(x+3\right)}=2+\dfrac{7}{x^2-9}\)
Để \(A\in Z\Leftrightarrow7⋮x^2-9\)
\(\Rightarrow x^2-9\in U\left(7\right)=\left\{1;7\right\}\left(x\in N\right)\)
\(\Rightarrow\left(x-3\right);\left(x+3\right)\in\left\{-1;-7;1;7\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}x-3=1\\x+3=7\end{matrix}\right.\) hay \(\left\{{}\begin{matrix}x-3=7\\x+3=1\end{matrix}\right.\) hay \(\left\{{}\begin{matrix}x-3=-1\\x+3=7\end{matrix}\right.\) hay \(\left\{{}\begin{matrix}x-3=7\\x+3=-1\end{matrix}\right.\)
\(\Rightarrow x\in\left\{4\right\}\)
Vậy \(x\in\left\{4\right\}\) thỏa mãn đề bài
em c.ơn trước
