$\sqrt{3}·\sqrt{12}$
= $\sqrt{3}·\sqrt{3·4}$
= $\sqrt{3}·\sqrt{3}·\sqrt{4}$
= $3·2$
= $6$

$\sqrt{\frac{5}{2}}·\sqrt{\frac{2}{5}}$
= $\sqrt{\frac{5}{2}·\frac{2}{5}}$
= $\sqrt{1·1}$
= $\sqrt{1}$
= $1$

$\sqrt{10}·\left(\sqrt{10}+\sqrt{160}\right)$
= $\sqrt{10}·\sqrt{10}+\sqrt{10}·\sqrt{160}$
= $10+\sqrt{10}·\sqrt{10·16}$
= $10+\sqrt{10}·\sqrt{10}·\sqrt{16}$
= $10+10·4$
= $10+40$
= $50$

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

${\left(\sqrt{3}-\sqrt{48}\right)}^2$ = ${\left(\sqrt{3}\right)}^2-2·\sqrt{3}·\sqrt{48}+{\left(\sqrt{48}\right)}^2$
= $3-2·\sqrt{3}·\sqrt{3·16}+48$
= $3-2·\sqrt{3}·\sqrt{3}·\sqrt{16}+48$
= $3-2·3·4+48$
= $3-24+48$
= $27$

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

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

$-10\sqrt{15}-9\sqrt{15}$
= $-19\sqrt{15}$