Answer
$\dfrac{\sqrt{3}+\sqrt{5}}{2}=-\dfrac{1}{\sqrt{3}-\sqrt{5}}$
Work Step by Step
$\dfrac{\sqrt{3}+\sqrt{5}}{2}$
Multiply both numerator and denominator by $\sqrt{3}-\sqrt{5}$ and simplify:
$\dfrac{\sqrt{3}+\sqrt{5}}{2}=\Big(\dfrac{\sqrt{3}+\sqrt{5}}{2}\Big)\Big(\dfrac{\sqrt{3}-\sqrt{5}}{\sqrt{3}-\sqrt{5}}\Big)=\dfrac{(\sqrt{3})^{2}-(\sqrt{5})^{2}}{2\sqrt{3}-2\sqrt{5}}=...$
$...=\dfrac{3-5}{2\sqrt{3}-2\sqrt{5}}=-\dfrac{2}{2(\sqrt{3}-\sqrt{5})}=-\dfrac{1}{\sqrt{3}-\sqrt{5}}$
