4. Zadatak

Prvi način:

S obzirom da su zadani izrazi pozitivni dokazom jednakosti njihovih kvadrata dokazujemo i njihovu jednakost.

$$\left(2\sqrt{21}\right)^2 = \left(\sqrt{22 + 2\sqrt{21}} + \sqrt{22 - 2\sqrt{21}}\right)^2.$$

Dobijemo:

$$84 = \left(\sqrt{22 + 2\sqrt{21}}\right)^2 + 2 \cdot \sqrt{22 + 2\sqrt{21}} \cdot \sqrt{22 - 2\sqrt{21}} + \left(\sqrt{22 - 2\sqrt{21}}\right)^2$$ $$84 = 22 + 2\sqrt{21} + 2\sqrt{(22 + 2\sqrt{21})\cdot(22 - 2\sqrt{21})} + 22 - 2\sqrt{21}$$ $$84 = 44 + 2\sqrt{22^2 - 4 \cdot 21}$$ $$84 = 44 + 2\sqrt{484 - 84}$$ $$84 = 44 + 2\sqrt{400}$$ $$84 = 84.$$

Drugi način:

$$\sqrt{22 + 2\sqrt{21}} + \sqrt{22 - 2\sqrt{21}} = \sqrt{21 + 2\sqrt{21} + 1} + \sqrt{21 - 2\sqrt{21} + 1}$$ $$= \sqrt{(\sqrt{21} + 1)^2} + \sqrt{(\sqrt{21} - 1)^2}.$$

Budući da je $\sqrt{21} + 1 > 0$ i $\sqrt{21} - 1 > 0$, vrijedi

$$\sqrt{(\sqrt{21} + 1)^2} + \sqrt{(\sqrt{21} - 1)^2} = \sqrt{21} + 1 + \sqrt{21} - 1 = 2\sqrt{21}.$$