1. Zadatak

Racionalizirajmo razlomke:

$$\frac{1}{\sqrt{15}+\sqrt{13}}\cdot\frac{\sqrt{15}-\sqrt{13}}{\sqrt{15}-\sqrt{13}}=\frac{\sqrt{15}-\sqrt{13}}{2},$$ $$\frac{1}{\sqrt{13}-\sqrt{11}}\cdot\frac{\sqrt{13}+\sqrt{11}}{\sqrt{13}+\sqrt{11}}=\frac{\sqrt{13}+\sqrt{11}}{2}.$$

Zadnji razlomak zapišimo malo drugačije:

$$\frac{\sqrt{5.5}-\sqrt{7.5}}{\sqrt{2}}=\frac{\sqrt{\frac{11}{2}}-\sqrt{\frac{15}{2}}}{\sqrt{2}}=\frac{\frac{\sqrt{11}-\sqrt{15}}{\sqrt{2}}}{\sqrt{2}}=\frac{\sqrt{11}-\sqrt{15}}{2}.$$

(Ili:

$$\frac{\sqrt{5.5}-\sqrt{7.5}}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{11}-\sqrt{15}}{2}.$$

)

Na kraju zbrojimo sve dobiveno:

$$\frac{1}{\sqrt{15}+\sqrt{13}}+\frac{1}{\sqrt{13}-\sqrt{11}}-\frac{\sqrt{5.5}-\sqrt{7.5}}{\sqrt{2}}=\frac{\sqrt{15}-\sqrt{13}}{2}+\frac{\sqrt{13}+\sqrt{11}}{2}-\frac{\sqrt{11}-\sqrt{15}}{2}=\sqrt{15}.$$