$\frac{\sqrt{12}}{\sqrt{3}}$
= $\sqrt{\frac{12}{3}}$
= $\sqrt{4}$
= 2

$\frac{\sqrt{0.32}}{\sqrt{0.02}}$
= $\sqrt{\frac{0.32}{0.02}}$
= $\sqrt{\frac{32}{2}}$
= $\sqrt{16}$
= 4

$\sqrt{2}·\left(\sqrt{50}+\sqrt{2}\right)$
= $\sqrt{2}·\sqrt{50}+\sqrt{2}·\sqrt{2}$
= $\sqrt{2}·\sqrt{2·25}+2$
= $\sqrt{2}·\sqrt{2}·\sqrt{25}+2$
= $2·5+2$
= $10+2$
= $12$

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

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

$\sqrt{25x}-5\sqrt{x}$
= $\sqrt{25}·\sqrt{x}-5\sqrt{x}$
= $5\sqrt{x}-5\sqrt{x}$
= 0

Wir können die $\sqrt{35}$ wie Einheiten behandeln und einfach die $\sqrt{35}$ verrechnen.

$10\sqrt{35}+\sqrt{35}$
= $11\sqrt{35}$