$\frac{\sqrt{250}}{\sqrt{10}}$
= $\sqrt{\frac{250}{10}}$
= $\sqrt{25}$
= 5

$\frac{\sqrt{\frac{1}{15}}}{\sqrt{\frac{3}{5}}}$
= $\sqrt{\frac{\frac{1}{15}}{\frac{3}{5}}}$
= $\sqrt{\frac{1}{15}·\frac{5}{3}}$
= $\sqrt{\frac{1}{3}·\frac{1}{3}}$
= $\sqrt{\frac{1}{9}}$
= $\frac{1}{3}$

Wir können die $\sqrt{x}$ wie Einheiten behandeln und ausklammern, bzw. einfach verrechnen.

$-8\sqrt{x}-\sqrt{x}$

= $\left(-8-1\right)·\sqrt{x}$

= $-9\sqrt{x}$

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

Nach der 3. binomischen Formel ((a+b)(a-b)=a²-b²) gilt:

$\left(\sqrt{6}+\sqrt{5}\right)·\left(\sqrt{6}-\sqrt{5}\right)$ = ${\left(\sqrt{6}\right)}^2-{\left(\sqrt{5}\right)}^2$
= $6-5$
= $1$

$\frac{\sqrt{40x}}{\sqrt{10x}}$
= $\frac{\sqrt{40}·\sqrt{x}}{\sqrt{10}·\sqrt{x}}$
= $\frac{\sqrt{40}}{\sqrt{10}}$
= $\sqrt{\frac{40}{10}}$
= $\sqrt{4}$
= $2$

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

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