Монгол Бодлогын Сан

Заавар: $\alpha+\beta=\gamma$ бол $$\begin{align*} (\sqrt\alpha+\sqrt\beta&{}+\sqrt\gamma)(\sqrt\alpha+\sqrt\beta-\sqrt\gamma)=\\ &=(\sqrt\alpha+\sqrt\beta)^2-(\sqrt\gamma)^2\\ &=\alpha+2\sqrt\alpha\sqrt\beta+\beta-\gamma\\ &=2\sqrt{\alpha\beta} \end{align*}$$

Бодолт: $$\begin{align*} \alpha&=\dfrac{1}{\sqrt{11}+\sqrt{5}+\sqrt{6}}=\\ &=\dfrac{(\sqrt{5}+\sqrt{6}-\sqrt{11})}{(\sqrt{5}+\sqrt{6}+\sqrt{11})(\sqrt{5}+\sqrt{6}-\sqrt{11})}\\ &=\dfrac{\sqrt{5}+\sqrt{6}-\sqrt{11}}{2\sqrt5\sqrt6} & \color{red}{\dfrac{{}\times\sqrt5\sqrt6}{{}\times\sqrt5\sqrt6}}\\ &=\dfrac{6\sqrt5+5\sqrt6-\sqrt{330}}{60} \end{align*}$$