\(\sqrt{x^2-8x-9}=\sqrt{\left(x-9\right)\left(x+1\right)}\)
ĐKXĐ: \(\left(x-9\right)\left(x+1\right)\ge0\Leftrightarrow\left[{}\begin{matrix}x\le-1\\x\ge9\end{matrix}\right.\)
\(\sqrt{x^2-8x-9}\\ =\sqrt{x^2-8x+16-25}\\=\sqrt{\left(x^2-8x+16\right)-25}\\ =\sqrt{\left(x-4\right)^2-5^2}\\ =\sqrt{\left(x-9\right)\left(x+1\right)}\\ =\sqrt{x-9}\sqrt{x+1}\)
ĐKXĐ: \(\left[{}\begin{matrix}x\ge9\\x\le-1\end{matrix}\right.\)