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

$\frac{\sqrt{\frac{1}{5}}}{\sqrt{\frac{4}{5}}}$
= $\sqrt{\frac{\frac{1}{5}}{\frac{4}{5}}}$
= $\sqrt{\frac{1}{5}·\frac{5}{4}}$
= $\sqrt{1·\frac{1}{4}}$
= $\sqrt{\frac{1}{4}}$
= $\frac{1}{2}$

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

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

$\left(\sqrt{7}+\sqrt{6}\right)·\left(\sqrt{7}-\sqrt{6}\right)$ = ${\left(\sqrt{7}\right)}^2-{\left(\sqrt{6}\right)}^2$
= $7-6$
= $1$

$\sqrt{3x}·\sqrt{27x}$
= $\sqrt{3}·\sqrt{x}·\sqrt{27}·\sqrt{x}$
= $\sqrt{3}·\sqrt{27}·\sqrt{x}·\sqrt{x}$
= $\sqrt{3}·\sqrt{3·9}·x$
= $\sqrt{3}·\sqrt{3}·\sqrt{9}·x$
= $3·3·x$
= $9x$

Wir können die $\sqrt{45}$ wie Einheiten behandeln und einfach verrechnen.

$6\sqrt{45}-3\sqrt{45}$
= $3\sqrt{45}$