1. Zadatak

Prvo rješenje.

Računamo redom:

$$\begin{aligned} (\sqrt{666}+\sqrt{888})^2&=\\ 666+2\cdot\sqrt{666}\cdot\sqrt{888}+888&=\\ 1554+2\cdot\sqrt6\cdot\sqrt{111}\cdot\sqrt8\cdot\sqrt{111}&=\\ 1554+222\sqrt{48}&=\\ 1554+222\cdot4\sqrt3&=\\ 1554+888\sqrt3 \end{aligned}$$ $$\begin{aligned} \sqrt{666^2+888^2}&=\\ \sqrt{6^2\cdot111^2+8^2\cdot111^2}&=\\ \sqrt{111^2\cdot(6^2+8^2)}&=\\ 111\cdot\sqrt{100}&=\\ 1110 \end{aligned}$$

Konačno:

$$\begin{aligned} \frac{(\sqrt{666}+\sqrt{888})^2-\sqrt{666^2+888^2}}{444}&=\\ \frac{1554+888\sqrt3-1110}{444}&=\\ \frac{444+888\sqrt3}{444}&=\\ 1+2\sqrt3. \end{aligned}$$

Drugo rješenje.

$$\begin{aligned} &\frac{(\sqrt{666}+\sqrt{888})^2-\sqrt{666^2+888^2}}{444}\\ &=\frac{(\sqrt{111\cdot6}+\sqrt{111\cdot8})^2-\sqrt{(111\cdot6)^2+(111\cdot8)^2}}{111\cdot4}\\ &=\frac{(\sqrt{111}\cdot\sqrt6+\sqrt{111}\cdot\sqrt8)^2-\sqrt{111^2\cdot6^2+111^2\cdot8^2}}{111\cdot4}\\ &=\frac{(\sqrt{111}\cdot(\sqrt6+\sqrt8))^2-\sqrt{111^2\cdot(6^2+8^2)}}{111\cdot4}\\ &=\frac{\sqrt{111}^{,2}\cdot(\sqrt6+\sqrt8)^2-\sqrt{111^2\cdot(36+64)}}{111\cdot4}\\ &=\frac{111\cdot(6+2\sqrt6\cdot\sqrt8+8)-111\sqrt{100}}{111\cdot4}\\ &=\frac{111\cdot(14+2\sqrt{48})-111\cdot10}{111\cdot4}\\ &=\frac{111\cdot(14+2\sqrt{16}\cdot3-10)}{111\cdot4}\\ &=\frac{4+8\sqrt3}{4}\\ &=1+2\sqrt3. \end{aligned}$$