Монгол Бодлогын Сан

Заавар: $$\dfrac{1}{a\pm\sqrt{b}}=\dfrac{a\mp\sqrt{b}}{a^2-b}$$

Бодолт: $$\begin{align*} \text{Илэрх.}&=\left(\dfrac2{\sqrt3-1}+\dfrac3{\sqrt3-2}+\dfrac{15}{3-\sqrt3}\right)\cdot(\sqrt3+5)^{-1}\\ &=\left(\dfrac{2(\sqrt3+1)}{3-1}+\dfrac{3(\sqrt3+2)}{3-4}+\dfrac{15(3+\sqrt3)}{9-3}\right)\cdot\dfrac{1}{\sqrt{3}+5}\\ &=\left(\sqrt3+1-3\sqrt3-6+\dfrac{15+5\sqrt3}{2}\right)\cdot\dfrac{1}{\sqrt{3}+5}\\ &=\dfrac{5+\sqrt3}{2}\cdot\dfrac{1}{\sqrt3+5}=\dfrac12 \end{align*}$$