1. Zadatak

Prvo rješenje.

$$\begin{aligned} &\frac{1}{\sqrt{2}-1}-\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}+1}\\ &=\frac{\sqrt{2}+1-(\sqrt{2}-1)}{(\sqrt{2}-1)(\sqrt{2}+1)} +\frac{\sqrt{3}+1-(\sqrt{3}-1)}{(\sqrt{3}-1)(\sqrt{3}+1)}\\ &=\frac{\sqrt{2}+1-\sqrt{2}+1}{(\sqrt{2})^2-1^2} +\frac{\sqrt{3}+1-\sqrt{3}+1}{(\sqrt{3})^2-1^2}\\ &=\frac{2}{2-1}+\frac{2}{3-1}=2+1=3. \end{aligned}$$

Drugo rješenje.

$$\begin{aligned} &\frac{1}{\sqrt{2}-1}-\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}+1}\\ &=\frac{1}{\sqrt{2}-1}\cdot\frac{\sqrt{2}+1}{\sqrt{2}+1} -\frac{1}{\sqrt{2}+1}\cdot\frac{\sqrt{2}-1}{\sqrt{2}-1}\\ &\quad+\frac{1}{\sqrt{3}-1}\cdot\frac{\sqrt{3}+1}{\sqrt{3}+1} -\frac{1}{\sqrt{3}+1}\cdot\frac{\sqrt{3}-1}{\sqrt{3}-1}\\ &=\frac{\sqrt{2}+1}{(\sqrt{2})^2-1^2} -\frac{\sqrt{2}-1}{(\sqrt{2})^2-1^2} +\frac{\sqrt{3}+1}{(\sqrt{3})^2-1^2} -\frac{\sqrt{3}-1}{(\sqrt{3})^2-1^2}\\ &=\frac{\sqrt{2}+1-\sqrt{2}+1}{(\sqrt{2})^2-1^2} +\frac{\sqrt{3}+1-\sqrt{3}+1}{(\sqrt{3})^2-1^2}\\ &=\frac{2}{2-1}+\frac{2}{3-1}=2+1=3. \end{aligned}$$