$\sqrt{250}·\sqrt{10}$
= $\sqrt{10·25}·\sqrt{10}$
= $\sqrt{25}·\sqrt{10}·\sqrt{10}$
= $5·10$
= $50$

$\frac{\sqrt{\mathrm{0,8}}}{\sqrt{\mathrm{0,2}}}$
= $\sqrt{\frac{0.8}{0.2}}$
= $\sqrt{\frac{8}{2}}$
= $\sqrt{4}$
= 2

$\left(\sqrt{5}+\sqrt{80}\right)·\sqrt{5}$
= $\sqrt{5}·\sqrt{5}+\sqrt{80}·\sqrt{5}$
= $5+\sqrt{5·16}·\sqrt{5}$
= $5+\sqrt{16}·\sqrt{5}·\sqrt{5}$
= $5+4·5$
= $5+20$
= $25$

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

${\left(\sqrt{10}+\sqrt{90}\right)}^2$ = ${\left(\sqrt{10}\right)}^2+2·\sqrt{10}·\sqrt{90}+{\left(\sqrt{90}\right)}^2$
= $10+2·\sqrt{10}·\sqrt{10·9}+90$
= $10+2·\sqrt{10}·\sqrt{10}·\sqrt{9}+90$
= $10+2·10·3+90$
= $10+60+90$
= $160$

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

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

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