\(ĐK:1-x^2>0\Leftrightarrow\left(1-x\right)\left(1+x\right)>0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}1-x>0\\1+x>0\end{matrix}\right.\\\left\{{}\begin{matrix}1-x< 0\\1+x< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 1\\x>-1\end{matrix}\right.\\\left\{{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-1< x< 1\\x\in\varnothing\end{matrix}\right.\\ \Leftrightarrow-1< x< 1\)
\(\dfrac{1}{\sqrt{1-x^2}}\) có đkxđ khi 1 - x2 > 0
<=> (1 - x)(1 + x) > 0
<=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}1-x>0\\1+x< 0\end{matrix}\right.\\\left[{}\begin{matrix}1-x< 0\\1+x>0\end{matrix}\right.\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}x< 1\\x< -1\end{matrix}\right.\\\left[{}\begin{matrix}x>1\\x>-1\end{matrix}\right.\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x>1\\x>1\end{matrix}\right.\)
Vậy ĐKXĐ là x > 1
