Обчисліть:
1) $(\sqrt{3+\sqrt5}+\sqrt{3-\sqrt5})^2;$
2) $\frac{2}{9+4\sqrt5}+\frac{2}{9-4\sqrt5}.$
Розв'язок:
1) $(\sqrt{3+\sqrt5}+\sqrt{3-\sqrt5})^2=$
$=(\sqrt{3+\sqrt5})^2+$
$+ 2\sqrt{(3+\sqrt5)(3-\sqrt5)}+$
$+ (\sqrt{3-\sqrt5})^2=$
$=3+\sqrt5+2\sqrt{3^2-(\sqrt5)^2}+$
$+ 3-\sqrt5=6+2\sqrt{9-5}=$
$= 6+2\sqrt4=6+4=10.$
2) $\frac{2}{9+4\sqrt5}+\frac{2}{9-4\sqrt5}=$
$= \frac{2(9-4\sqrt5)+2(9+4\sqrt5)}{(9+4\sqrt5)(9-4\sqrt5)}=$
$=\frac{18-8\sqrt5+18+8\sqrt5}{9^2-(4\sqrt5)^2}=\frac{36}{81-80}=$
$= \frac{36}{1}=36.$