Монгол Бодлогын Сан

Заавар: $\Bigl(\dfrac{1}{\sqrt{3}+2}^{\color{red}{(\sqrt{3}-2}}-\dfrac{1}{\sqrt{3}-2}^{\color{red}{(\sqrt{3}+2}}\Bigr):\Bigl(1^{\color{red}{(\sqrt3}}+\dfrac{2}{\sqrt{3}}\Bigr)$ гээд бод. $$\dfrac{a}{b}:\dfrac{c}{d}=\dfrac{a}{b}\cdot\dfrac{d}{c}$$ гэдгийг санаарай!

Бодолт: \begin{align*} \Bigl(\dfrac{1}{\sqrt{3}+2}^{\color{red}{(\sqrt{3}-2}}&-\dfrac{1}{\sqrt{3}-2}^{\color{red}{(\sqrt{3}+2}}\Bigr):\Bigl(1^{\color{red}{(\sqrt3}}+\dfrac{2}{\sqrt{3}}\Bigr)=\\ &=\left(\dfrac{(\sqrt3-2)-(\sqrt3+2)}{(\sqrt3+2)(\sqrt3-2)}\right):\left(\dfrac{\sqrt3+2}{\sqrt3}\right)=\\ &=\left(\dfrac{-4}{(\sqrt3)^2-2^2}\right)\cdot\left(\dfrac{\sqrt3}{\sqrt3+2}\right)=\dfrac{4\sqrt3}{\sqrt3+2} \end{align*}