a) ta có : \(C=\dfrac{x-3}{x+6}=\dfrac{x+6-9}{x+6}=1-\dfrac{9}{x+6}\) là phân số
\(\Leftrightarrow\dfrac{9}{x+6}\) là số phân số \(\Leftrightarrow x+6\ne\) ước của 9 là \(\pm1;\pm3;\pm9\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+6\ne1\\x+6\ne-1\end{matrix}\right.\\\left\{{}\begin{matrix}x+6\ne3\\x+6\ne-3\end{matrix}\right.\\\left\{{}\begin{matrix}x+6\ne9\\x+6\ne-9\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ne-5\\x\ne-7\end{matrix}\right.\\\left\{{}\begin{matrix}x\ne-3\\x\ne-9\end{matrix}\right.\\\left\{{}\begin{matrix}x\ne3\\x\ne-15\end{matrix}\right.\end{matrix}\right.\) vậy .........................................
b) ta có : \(C=\dfrac{x-3}{x+6}=\dfrac{x+6-9}{x+6}=1-\dfrac{9}{x+6}\) nguyên
\(\Leftrightarrow\dfrac{9}{x+6}\) nguyên \(\Leftrightarrow x+6\) thuộc ước của 9 là \(\pm1;\pm2;\pm3\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+6=1\\x+6=-1\end{matrix}\right.\\\left[{}\begin{matrix}x+6=3\\x+6=-3\end{matrix}\right.\\\left[{}\begin{matrix}x+6=9\\x+6=-9\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=-5\\x=-7\end{matrix}\right.\\\left[{}\begin{matrix}x=-3\\x=-9\end{matrix}\right.\\\left[{}\begin{matrix}x=3\\x=-15\end{matrix}\right.\end{matrix}\right.\) vậy ..............................................
