Монгол Бодлогын Сан

Заавар: $$6\pm\sqrt2=(2\pm\sqrt2)^2$$

Бодолт: $$\begin{align*} \text{Илэрх.}&=\dfrac{4}{\sqrt{2}-\sqrt{6+4\sqrt{2}}}+\dfrac{4}{\sqrt{2}+ \sqrt{6-4\sqrt{2}}}\\ &=\dfrac{4}{\sqrt2-\sqrt{(2+\sqrt2)^2}}+\dfrac{4}{\sqrt2+\sqrt{(2-\sqrt2)^2}}\\ &=\dfrac{4}{\sqrt2-|2+\sqrt2|}+\dfrac{4}{\sqrt2+|2-\sqrt2|}\\ &=\dfrac{4}{\sqrt2-2-\sqrt2}-\dfrac{4}{\sqrt2+2-\sqrt2}\\ &=-2+2=0 \end{align*}$$