1. Zadatak

Uočimo da vrijedi:

$$\sqrt{6+4\sqrt2}=\sqrt{4+4\sqrt2+2}=\sqrt{(2+\sqrt2)^2}=|2+\sqrt2|=2+\sqrt2$$ $$\sqrt{6-4\sqrt2}=\sqrt{4-4\sqrt2+2}=\sqrt{(2-\sqrt2)^2}=|2-\sqrt2|=2-\sqrt2.$$

Uvrštavanjem dobivenih jednakosti u početni brojevni izraz odredimo njegovu vrijednost.

$$\begin{aligned} &\left(\frac{6+4\sqrt2}{\sqrt2+\sqrt{6+4\sqrt2}}+\frac{6-4\sqrt2}{\sqrt2-\sqrt{6-4\sqrt2}}\right)^2\\ &=\left(\frac{6+4\sqrt2}{\sqrt2+2+\sqrt2}+\frac{6-4\sqrt2}{\sqrt2-2+\sqrt2}\right)^2\\ &=\left(\frac{6+4\sqrt2}{2\sqrt2+2}\cdot\frac{2\sqrt2-2}{2\sqrt2-2}+\frac{6-4\sqrt2}{2\sqrt2-2}\cdot\frac{2\sqrt2+2}{2\sqrt2+2}\right)^2\\ &=\left(\frac{12\sqrt2+16-12-8\sqrt2}{(2\sqrt2)^2-2^2}+\frac{12\sqrt2-16+12-8\sqrt2}{(2\sqrt2)^2-2^2}\right)^2\\ &=\left(\frac{4\sqrt2+4}{4}+\frac{4\sqrt2-4}{4}\right)^2\\ &=(\sqrt2+1+\sqrt2-1)^2\\ &=(2\sqrt2)^2=8. \end{aligned}$$