Answer
$\dfrac{1}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{x-1}$
Work Step by Step
$\dfrac{1}{\sqrt{x}+1}$
Multiply both numerator and denominator by $-\sqrt{x}+1$ and simplify:
$\dfrac{1}{\sqrt{x}+1}=\Big(\dfrac{1}{\sqrt{x}+1}\Big)\Big(\dfrac{-\sqrt{x}+1}{-\sqrt{x}+1}\Big)=\dfrac{1-\sqrt{x}}{1-x}=\dfrac{\sqrt{x}-1}{x-1}$
