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

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

$\left(\sqrt{12}+\sqrt{3}\right)·\sqrt{3}$
= $\sqrt{12}·\sqrt{3}+\sqrt{3}·\sqrt{3}$
= $\sqrt{3·4}·\sqrt{3}+3$
= $\sqrt{4}·\sqrt{3}·\sqrt{3}+3$
= $2·3+3$
= $6+3$
= $9$

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

${\left(\sqrt{2}-\sqrt{72}\right)}^2$ = ${\left(\sqrt{2}\right)}^2-2·\sqrt{2}·\sqrt{72}+{\left(\sqrt{72}\right)}^2$
= $2-2·\sqrt{2}·\sqrt{2·36}+72$
= $2-2·\sqrt{2}·\sqrt{2}·\sqrt{36}+72$
= $2-2·2·6+72$
= $2-24+72$
= $50$

$\sqrt{80x}·\sqrt{5x}$
= $\sqrt{80}·\sqrt{x}·\sqrt{5}·\sqrt{x}$
= $\sqrt{80}·\sqrt{5}·\sqrt{x}·\sqrt{x}$
= $\sqrt{5·16}·\sqrt{5}·x$
= $\sqrt{16}·\sqrt{5}·\sqrt{5}·x$
= $4·5·x$
= $20x$

Wir können die $\sqrt{15}$ wie Einheiten behandeln und einfach die $\sqrt{15}$ verrechnen.

$7\sqrt{15}+6\sqrt{15}$
= $13\sqrt{15}$