Монгол Бодлогын Сан

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

Бодолт:\begin{align*} \text{Илэрх.}&=\dfrac{1}{\sqrt{x+y}-\sqrt{x}}-\dfrac{1}{\sqrt{x+y}+\sqrt{x}}\\ &=\dfrac{(\sqrt{x+y}+\sqrt{x})}{(\sqrt{x+y}-\sqrt{x})(\sqrt{x+y}+\sqrt{x})}-\dfrac{(\sqrt{x+y}-\sqrt{x})}{(\sqrt{x+y}+\sqrt{x})(\sqrt{x+y}-\sqrt{x})}\\ &=\dfrac{\sqrt{x+y}+\sqrt{x}}{(\sqrt{x+y})^2-(\sqrt{x})^2}-\dfrac{\sqrt{x+y}-\sqrt{x}}{(\sqrt{x+y})^2-(\sqrt{x})^2}\\ &=\dfrac{\sqrt{x+y}+\sqrt{x}}{x+y-x}-\dfrac{\sqrt{x+y}-\sqrt{x}}{x+y-x}=\dfrac{\sqrt{x+y}+\sqrt{x}}{y}-\dfrac{\sqrt{x+y}-\sqrt{x}}{y}\\ &=\dfrac{(\sqrt{x+y}+\sqrt{x})-(\sqrt{x+y}-\sqrt{x})}{y}=\dfrac{2\sqrt{x}}{y} \end{align*}