Монгол Бодлогын Сан

Заавар: $(a-\sqrt{2b})(a+\sqrt{2b})=a^2-(\sqrt{2b})^2=a^2-2b$ байна.

Бодолт: $$\left(\dfrac{1}{a-\sqrt{2b}}+\dfrac{1}{a+\sqrt{2b}}\right):\dfrac{a}{a^2-2b}=$$ $$=\left(\dfrac{(a+\sqrt{2b})}{(a-\sqrt{2b})(a+\sqrt{2b})}+\dfrac{(a-\sqrt{2b})}{(a-\sqrt{2b})(a+\sqrt{2b})}\right)\cdot\dfrac{a^2-2b}{a}=$$ $$=\dfrac{2a}{a^2-2b}\cdot \dfrac{a^2-2b}{a}=2.$$