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

$\sqrt{\frac{1}{125}}·\sqrt{5}$
= $\sqrt{\frac{1}{125}·5}$
= $\sqrt{\frac{1}{25}·1}$
= $\sqrt{\frac{1}{25}}$
= $\frac{1}{5}$

$\left(\sqrt{8}+\sqrt{2}\right)·\sqrt{2}$
= $\sqrt{8}·\sqrt{2}+\sqrt{2}·\sqrt{2}$
= $\sqrt{2·4}·\sqrt{2}+2$
= $\sqrt{4}·\sqrt{2}·\sqrt{2}+2$
= $2·2+2$
= $4+2$
= $6$

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

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

$\sqrt{48x}·\sqrt{3x}$
= $\sqrt{48}·\sqrt{x}·\sqrt{3}·\sqrt{x}$
= $\sqrt{48}·\sqrt{3}·\sqrt{x}·\sqrt{x}$
= $\sqrt{3·16}·\sqrt{3}·x$
= $\sqrt{16}·\sqrt{3}·\sqrt{3}·x$
= $4·3·x$
= $12x$

Wir können die $\sqrt{2}$ wie Einheiten behandeln und einfach verrechnen.

$-13\sqrt{2}-6\sqrt{2}$
= $-19\sqrt{2}$