Монгол Бодлогын Сан

Заавар: $$\dfrac{1}{a+\sqrt{b}}=\dfrac{a-\sqrt{b}}{a^2-b}$$ томьёог ашигла.

Бодолт: $a=\sqrt2+\sqrt5$, $b=10$ гээд зааварт байгаа томьёог ашиглавал $$\begin{align*} \dfrac{1}{\sqrt{2}+\sqrt{5}+\sqrt{10}}&=\dfrac{\sqrt{2}+\sqrt{5}-\sqrt{10}}{(\sqrt2+\sqrt5)^2-10}=\dfrac{\sqrt{2}+\sqrt{5}-\sqrt{10}}{7+2\sqrt{10}-10}\\ &=\dfrac{(\sqrt{2}+\sqrt{5}-\sqrt{10})(2\sqrt{10}+3)}{(2\sqrt{10}-3)(2\sqrt{10}+3)}\\ &=\dfrac{\sqrt{2}(2\sqrt{10}+3)+\sqrt{5}(2\sqrt{10}+3)-\sqrt{10}(2\sqrt{10}+3)}{(2\sqrt{10}-3)(2\sqrt{10}+3)}\\ &=\dfrac{4\sqrt{5}+3\sqrt{2}+10\sqrt{2}+3\sqrt{5}-20-3\sqrt{10}}{(2\sqrt{10})^2-3^2}\\ &=\dfrac{13\sqrt{2}+7\sqrt{5}-3\sqrt{10}-20}{31} \end{align*}$$