Монгол Бодлогын Сан

Заавар: $$\begin{gather} \overline{0.(ab)}=\dfrac{\overline{ab}}{99} \end{gather}$$

Бодолт: $$\begin{gather} \dfrac{6}{11}=0.(54),\\ \dfrac{3(1+\sqrt3)}{3+\sqrt3}=\dfrac{3(1+\sqrt3)}{\sqrt3(1+\sqrt3)}=\sqrt3,\\ \dfrac{1}{\sqrt2-\sqrt3}=-(\sqrt2+\sqrt3),\\ \sqrt{5+2\sqrt{6}}=\sqrt2+\sqrt3 \end{gather}$$ болохыг тус тус тооцвол $$1.(54)-0.(54)+\sqrt3-(\sqrt2+\sqrt3)+\sqrt2+\sqrt3=1+\sqrt3.$$