a, Để \(\dfrac{x+3}{x-7}\) là số hữu tỉ thì: \(\left\{{}\begin{matrix}\dfrac{x+3}{x-7}\in\mathbb{Q}\\x-7\ne0\end{matrix}\right.\Rightarrow x\in\mathbb{Q};x\ne7\)
\(\Rightarrow\) \(\dfrac{x+3}{x-7}\) là số hữu tỉ dương khi: \(\dfrac{x+3}{x-7}>0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3>0\\x-7>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+3< 0\\x-7< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>-3\\x>7\end{matrix}\right.\\\left\{{}\begin{matrix}x< -3\\x< 7\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>7\\x< -3\end{matrix}\right.\)
b, Để \(\dfrac{x-10}{x-5}\) là số hữu tỉ thì: \(\left\{{}\begin{matrix}\dfrac{x-10}{x-5}\in\mathbb{Q}\\x-5\ne0\end{matrix}\right.\Rightarrow x\in\mathbb{Q};x\ne5\)
\(\Rightarrow\) \(\dfrac{x-10}{x-5}\) là số hữu tỉ âm khi: \(\dfrac{x-10}{x-5}< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-10< 0\\x-5>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-10>0\\x-5< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 10\\x>5\end{matrix}\right.\\\left\{{}\begin{matrix}x>10\\x< 5\end{matrix}\right.\left(\text{vô lí}\right)\end{matrix}\right.\)
\(\Rightarrow5< x< 10\)
#$\mathtt{Toru}$