$\frac{\sqrt{27}}{\sqrt{3}}$
= $\sqrt{\frac{27}{3}}$
= $\sqrt{9}$
= 3

$\sqrt{3}·\sqrt{\frac{1}{108}}$
= $\sqrt{3·\frac{1}{108}}$
= $\sqrt{1·\frac{1}{36}}$
= $\sqrt{\frac{1}{36}}$
= $\frac{1}{6}$

$\sqrt{2}·\left(\sqrt{32}+\sqrt{2}\right)$
= $\sqrt{2}·\sqrt{32}+\sqrt{2}·\sqrt{2}$
= $\sqrt{2}·\sqrt{2·16}+2$
= $\sqrt{2}·\sqrt{2}·\sqrt{16}+2$
= $2·4+2$
= $8+2$
= $10$

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

${\left(\sqrt{10}+\sqrt{160}\right)}^2$ = ${\left(\sqrt{10}\right)}^2+2·\sqrt{10}·\sqrt{160}+{\left(\sqrt{160}\right)}^2$
= $10+2·\sqrt{10}·\sqrt{10·16}+160$
= $10+2·\sqrt{10}·\sqrt{10}·\sqrt{16}+160$
= $10+2·10·4+160$
= $10+80+160$
= $250$

$\frac{\sqrt{12x}}{\sqrt{3x}}$
= $\frac{\sqrt{12}·\sqrt{x}}{\sqrt{3}·\sqrt{x}}$
= $\frac{\sqrt{12}}{\sqrt{3}}$
= $\sqrt{\frac{12}{3}}$
= $\sqrt{4}$
= $2$

Wir können die $\sqrt{5}$ wie Einheiten behandeln und einfach verrechnen.

$-\sqrt{5}-9\sqrt{5}$
= $-10\sqrt{5}$