$\frac{\sqrt{32}}{\sqrt{2}}$
= $\sqrt{\frac{32}{2}}$
= $\sqrt{16}$
= 4

$\sqrt{54}·\sqrt{\frac{1}{6}}$
= $\sqrt{54·\frac{1}{6}}$
= $\sqrt{9·1}$
= $\sqrt{9}$
= $3$

$\left(\sqrt{360}+\sqrt{10}\right)·\sqrt{10}$
= $\sqrt{360}·\sqrt{10}+\sqrt{10}·\sqrt{10}$
= $\sqrt{10·36}·\sqrt{10}+10$
= $\sqrt{36}·\sqrt{10}·\sqrt{10}+10$
= $6·10+10$
= $60+10$
= $70$

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

${\left(\sqrt{10}-\sqrt{90}\right)}^2$ = ${\left(\sqrt{10}\right)}^2-2·\sqrt{10}·\sqrt{90}+{\left(\sqrt{90}\right)}^2$
= $10-2·\sqrt{10}·\sqrt{10·9}+90$
= $10-2·\sqrt{10}·\sqrt{10}·\sqrt{9}+90$
= $10-2·10·3+90$
= $10-60+90$
= $40$

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

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

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