\(\dfrac{x}{x+2}+\sqrt{x-2}\) xác định \(\Leftrightarrow\left[{}\begin{matrix}x+2>0\\x-2\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x>-2\\x\ge2\end{matrix}\right.\)
\(\Leftrightarrow x\ge2\)
\(\dfrac{x}{x+2}+\sqrt{x-2}\)
Xác định khi:
\(\left\{{}\begin{matrix}x+2\ne0\\x-2\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-2\\x\ge2\end{matrix}\right.\)