ĐKXĐ
a) \(x\ge0\) và \(x\ne4\)
b)=\(\sqrt{\dfrac{x+2}{x-2}}\)
ĐKXĐ
TH1 \(\left\{{}\begin{matrix}X+2\ge0\\X-2>0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}X\ge-2\\X>2\end{matrix}\right.\) => X > 2
TH2 \(\left\{{}\begin{matrix}x+2\le0\\x-2< 0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\le-2\\x< 2\end{matrix}\right.\) => X \(\le\) -2
vậy ĐKXĐ\(\left[{}\begin{matrix}X>2\\X\le-2\end{matrix}\right.\)