1. Zadatak

Prvo rješenje.

$$\begin{aligned} &\left(\sqrt{7+2\sqrt{10}}+\sqrt{7-2\sqrt{10}}\right)^2\\ &=\left(\sqrt{7+2\sqrt{10}}+\sqrt{7-2\sqrt{10}}\right)\\ &\quad\cdot\left(\sqrt{7+2\sqrt{10}}+\sqrt{7-2\sqrt{10}}\right) \end{aligned}$$ $$\begin{aligned} &=\left(\sqrt{7+2\sqrt{10}}\right)^2 +\sqrt{(7+2\sqrt{10})(7-2\sqrt{10})}\\ &\quad+\sqrt{(7-2\sqrt{10})(7+2\sqrt{10})} +\left(\sqrt{7-2\sqrt{10}}\right)^2 \end{aligned}$$ $$\begin{aligned} &=\left(\sqrt{7+2\sqrt{10}}\right)^2 +2\cdot\sqrt{(7+2\sqrt{10})(7-2\sqrt{10})}\\ &\quad+\left(\sqrt{7-2\sqrt{10}}\right)^2 \end{aligned}$$ $$\begin{aligned} &=7+2\sqrt{10}\\ &\quad+2\cdot\sqrt{49-14\sqrt{10}+14\sqrt{10}-40}\\ &\quad+7-2\sqrt{10}\\ &=7+2\sqrt{10}+2\cdot\sqrt{49-40}+7-2\sqrt{10}\\ &=14+2\cdot3\\ &=20. \end{aligned}$$

Napomena: Učenik može, prilikom rješavanja zadatka, koristiti i izraze za kvadrat binoma te razliku kvadrata.

Drugo rješenje.

$$\begin{aligned} \left(\sqrt{7+2\sqrt{10}}+\sqrt{7-2\sqrt{10}}\right)^2 &=\left(\sqrt{(\sqrt2+\sqrt5)^2}+\sqrt{(\sqrt2-\sqrt5)^2}\right)^2\\ &=\left(\sqrt2+\sqrt5+\sqrt5-\sqrt2\right)^2\\ &=\left(2\sqrt5\right)^2\\ &=20. \end{aligned}$$