\(\dfrac{x}{x^2-4}+\sqrt{x-2}\)
\(=\dfrac{x}{\left(x-2\right)\left(x+2\right)}+\sqrt{x-2}\)
Ta thấy :
x - 2 < x + 2
\(\dfrac{x}{x^2-4}+\sqrt{x-2}\) có nghĩa khi
\(\Rightarrow\left\{{}\begin{matrix}x-2>0\\\sqrt{x-2}\ge0\end{matrix}\right.\)
\(\Leftrightarrow x>2\)