keeping in mind that
1² = 1
1³ = 1
1⁴ = 1
1¹⁰⁰⁰⁰⁰⁰⁰⁰⁰ = 1
then
[tex]\bf \begin{cases}
f(x)&=\sqrt{x^2-1}\\
g(x)&=\sqrt{x-1}\\
\left( \frac{f}{g} \right)(x)&=\frac{f(x)}{g(x)}
\end{cases}
\\\\\\
\cfrac{f(x)}{g(x)}\implies \cfrac{\sqrt{x^2-1}}{\sqrt{x-1}}\implies \sqrt{\cfrac{x^2-1}{x-1}}\implies \sqrt{\cfrac{x^2-1^2}{x-1}}
\\\\\\
\sqrt{\cfrac{\stackrel{\textit{difference of squares}}{\underline{(x-1)}(x+1)}}{\underline{x-1}}}\implies \sqrt{x+1}[/tex]
