Answer
$\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{x-1}{x-2\sqrt{x}+1}$
Work Step by Step
$\dfrac{\sqrt{x}+1}{\sqrt{x}-1}$
Multiply the numerator and the denominator of this expression by the conjugate of the numerator and simplify if possible:
$\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}-1}=\dfrac{(\sqrt{x})^{2}-1^{2}}{(\sqrt{x}-1)^{2}}=...$
$...=\dfrac{x-1}{(\sqrt{x})^{2}-2\sqrt{x}+1^{2}}=\dfrac{x-1}{x-2\sqrt{x}+1}$
