$\sqrt{5}·\sqrt{20}$
= $\sqrt{5}·\sqrt{5·4}$
= $\sqrt{5}·\sqrt{5}·\sqrt{4}$
= $5·2$
= $10$

$\sqrt{\mathrm{4,8}}·\sqrt{\mathrm{0,003}}$
= $\sqrt{\mathrm{4,8}·\mathrm{0,003}}$
= $\sqrt{\mathrm{0,0144}}$
= 0.12

$\sqrt{10}·\left(\sqrt{10}+\sqrt{250}\right)$
= $\sqrt{10}·\sqrt{10}+\sqrt{10}·\sqrt{250}$
= $10+\sqrt{10}·\sqrt{10·25}$
= $10+\sqrt{10}·\sqrt{10}·\sqrt{25}$
= $10+10·5$
= $10+50$
= $60$

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

${\left(\sqrt{5}+\sqrt{20}\right)}^2$ = ${\left(\sqrt{5}\right)}^2+2·\sqrt{5}·\sqrt{20}+{\left(\sqrt{20}\right)}^2$
= $5+2·\sqrt{5}·\sqrt{5·4}+20$
= $5+2·\sqrt{5}·\sqrt{5}·\sqrt{4}+20$
= $5+2·5·2+20$
= $5+20+20$
= $45$

$\sqrt{9x}-\sqrt{x}$
= $\sqrt{9}·\sqrt{x}-\sqrt{x}$
= $3\sqrt{x}-\sqrt{x}$
= $2\sqrt{x}$

Wir können die $\sqrt{27}$ wie Einheiten behandeln und einfach verrechnen.

$2\sqrt{27}-\sqrt{27}$
= $\sqrt{27}$