$\sqrt{8}·\sqrt{2}$
= $\sqrt{2·4}·\sqrt{2}$
= $\sqrt{4}·\sqrt{2}·\sqrt{2}$
= $2·2$
= $4$

$\frac{\sqrt{\mathrm{3,6}}}{\sqrt{\mathrm{0,1}}}$
= $\sqrt{\frac{3.6}{0.1}}$
= $\sqrt{36}$
= $\sqrt{36}$
= 6

Wir können die $\sqrt{x}$ wie Einheiten behandeln und ausklammern, bzw. einfach verrechnen.

$\sqrt{x}+3\sqrt{x}$

= $\left(1+3\right)·\sqrt{x}$

= $4\sqrt{x}$

$\sqrt{10}·\left(\sqrt{10}+\sqrt{360}\right)$
= $\sqrt{10}·\sqrt{10}+\sqrt{10}·\sqrt{360}$
= $10+\sqrt{10}·\sqrt{10·36}$
= $10+\sqrt{10}·\sqrt{10}·\sqrt{36}$
= $10+10·6$
= $10+60$
= $70$

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

${\left(\sqrt{10}-\sqrt{40}\right)}^2$ = ${\left(\sqrt{10}\right)}^2-2·\sqrt{10}·\sqrt{40}+{\left(\sqrt{40}\right)}^2$
= $10-2·\sqrt{10}·\sqrt{10·4}+40$
= $10-2·\sqrt{10}·\sqrt{10}·\sqrt{4}+40$
= $10-2·10·2+40$
= $10-40+40$
= $10$

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

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

${\left(\sqrt{108}+\sqrt{3}\right)}^2$ = ${\left(\sqrt{108}\right)}^2+2·\sqrt{108}·\sqrt{3}+{\left(\sqrt{3}\right)}^2$
= $108+2·\sqrt{3·36}·\sqrt{3}+3$
= $108+2·\sqrt{36}·\sqrt{3}·\sqrt{3}+3$
= $108+2·6·3+3$
= $108+36+3$
= $147$