$\frac{\sqrt{108}}{\sqrt{3}}$
= $\sqrt{\frac{108}{3}}$
= $\sqrt{36}$
= 6

$\frac{\sqrt{\frac{4}{15}}}{\sqrt{\frac{3}{5}}}$
= $\sqrt{\frac{\frac{4}{15}}{\frac{3}{5}}}$
= $\sqrt{\frac{4}{15}·\frac{5}{3}}$
= $\sqrt{\frac{4}{3}·\frac{1}{3}}$
= $\sqrt{\frac{4}{9}}$
= $\frac{2}{3}$

$\left(\sqrt{75}+\sqrt{3}\right)·\sqrt{3}$
= $\sqrt{75}·\sqrt{3}+\sqrt{3}·\sqrt{3}$
= $\sqrt{3·25}·\sqrt{3}+3$
= $\sqrt{25}·\sqrt{3}·\sqrt{3}+3$
= $5·3+3$
= $15+3$
= $18$

Nach der 2. binomischen Formel ((a-b)²=a²-2ab+b²) gilt:

${\left(\sqrt{3}-\sqrt{108}\right)}^2$ = ${\left(\sqrt{3}\right)}^2-2·\sqrt{3}·\sqrt{108}+{\left(\sqrt{108}\right)}^2$
= $3-2·\sqrt{3}·\sqrt{3·36}+108$
= $3-2·\sqrt{3}·\sqrt{3}·\sqrt{36}+108$
= $3-2·3·6+108$
= $3-36+108$
= $75$

$\sqrt{75x}·\sqrt{3x}$
= $\sqrt{75}·\sqrt{x}·\sqrt{3}·\sqrt{x}$
= $\sqrt{75}·\sqrt{3}·\sqrt{x}·\sqrt{x}$
= $\sqrt{3·25}·\sqrt{3}·x$
= $\sqrt{25}·\sqrt{3}·\sqrt{3}·x$
= $5·3·x$
= $15x$

Nach der 1. binomischen Formel ((a+b)²=a²+2ab+b²) gilt:

${\left(\sqrt{27}+\sqrt{3}\right)}^2$ = ${\left(\sqrt{27}\right)}^2+2·\sqrt{27}·\sqrt{3}+{\left(\sqrt{3}\right)}^2$
= $27+2·\sqrt{3·9}·\sqrt{3}+3$
= $27+2·\sqrt{9}·\sqrt{3}·\sqrt{3}+3$
= $27+2·3·3+3$
= $27+18+3$
= $48$