Только номера, отмеченные галочкой.

Решите задачу:

$\displaystyle 1) \frac{\left(3\sqrt{6}\right)^2}{18}=\frac{3^2\cdot \:6}{18}=\frac{3^2}{3}=3 \\ \\ \\ 3)\sqrt{3\frac{6}{7}}\cdot \sqrt{2\frac{1}{3}}=\sqrt{\frac{27}{7}}\sqrt{2 \frac{1}{3}}=\sqrt{\frac{27}{7}}\sqrt{\frac{7}{3}}=\frac{3\sqrt{3}}{\sqrt{7}}\sqrt{\frac{7}{3}}=\frac{3\sqrt{3}\sqrt{\frac{7}{3}}}{\sqrt{7}}=$
$=\displaystyle \frac{3\sqrt{3}\frac{\sqrt{7}}{\sqrt{3}}}{\sqrt{7}}=\frac{3\sqrt{7}}{\sqrt{7}}=3 \\ \\ \\ 5)\sqrt{16-\sqrt{31}}\sqrt{\sqrt{31}+16}=\sqrt{\left(16\sqrt{31}\right)\left(16+\sqrt{31}\right)}=$
$=6^2-\left(\sqrt{31}\right)^2=256-31=\sqrt{225}=15 \\ \\ \\ \displaystyle 7)\frac{\sqrt{3+\sqrt{5}}}{\sqrt{3-\sqrt{5}}}+\frac{\sqrt{3-\sqrt{5}}}{\sqrt{3+\sqrt{5}}}=\sqrt{\frac{3+\sqrt{5}}{3-\sqrt{5}}}+\sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}}=$
$\displaystyle = \sqrt{\frac{3\sqrt{5}+7}{2}}+\sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}}=\sqrt{\frac{3\sqrt{5}+7}{2}}+\sqrt{\frac{7-3\sqrt{5}}{2}}=3$