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

$\sqrt{3}·\sqrt{\frac{1}{75}}$
= $\sqrt{3·\frac{1}{75}}$
= $\sqrt{1·\frac{1}{25}}$
= $\sqrt{\frac{1}{25}}$
= $\frac{1}{5}$

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

$8\sqrt{x}-2\sqrt{x}$

= $\left(8-2\right)·\sqrt{x}$

= $6\sqrt{x}$

$\left(\sqrt{75}+\sqrt{3}\right)·\sqrt{3}$
= $\sqrt{75}·\sqrt{3}+\sqrt{3}·\sqrt{3}$
= $\sqrt{3·25}·\sqrt{3}+3$
= $\sqrt{25}·\sqrt{3}·\sqrt{3}+3$
= $5·3+3$
= $15+3$
= $18$

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

${\left(\sqrt{10}-\sqrt{160}\right)}^2$ = ${\left(\sqrt{10}\right)}^2-2·\sqrt{10}·\sqrt{160}+{\left(\sqrt{160}\right)}^2$
= $10-2·\sqrt{10}·\sqrt{10·16}+160$
= $10-2·\sqrt{10}·\sqrt{10}·\sqrt{16}+160$
= $10-2·10·4+160$
= $10-80+160$
= $90$

$\frac{\sqrt{360x}}{\sqrt{10x}}$
= $\frac{\sqrt{360}·\sqrt{x}}{\sqrt{10}·\sqrt{x}}$
= $\frac{\sqrt{360}}{\sqrt{10}}$
= $\sqrt{\frac{360}{10}}$
= $\sqrt{36}$
= $6$

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

${\left(\sqrt{12}-\sqrt{3}\right)}^2$ = ${\left(\sqrt{12}\right)}^2-2·\sqrt{12}·\sqrt{3}+{\left(\sqrt{3}\right)}^2$
= $12-2·\sqrt{3·4}·\sqrt{3}+3$
= $12-2·\sqrt{4}·\sqrt{3}·\sqrt{3}+3$
= $12-2·2·3+3$
= $12-12+3$
= $3$