$\frac{\sqrt{48}}{\sqrt{3}}$
= $\sqrt{\frac{48}{3}}$
= $\sqrt{16}$
= 4

$\frac{\sqrt{\frac{1}{10}}}{\sqrt{\frac{8}{5}}}$
= $\sqrt{\frac{\frac{1}{10}}{\frac{8}{5}}}$
= $\sqrt{\frac{1}{10}·\frac{5}{8}}$
= $\sqrt{\frac{1}{2}·\frac{1}{8}}$
= $\sqrt{\frac{1}{16}}$
= $\frac{1}{4}$

Wir können die $\sqrt{41}$ wie Einheiten behandeln und ausklammern, bzw. einfach verrechnen.

$-8\sqrt{41}-2\sqrt{41}$

= $\left(-8-2\right)·\sqrt{41}$

= $-10\sqrt{41}$

$\sqrt{3}·\left(\sqrt{12}+\sqrt{3}\right)$
= $\sqrt{3}·\sqrt{12}+\sqrt{3}·\sqrt{3}$
= $\sqrt{3}·\sqrt{3·4}+3$
= $\sqrt{3}·\sqrt{3}·\sqrt{4}+3$
= $3·2+3$
= $6+3$
= $9$

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

${\left(\sqrt{2}+\sqrt{8}\right)}^2$ = ${\left(\sqrt{2}\right)}^2+2·\sqrt{2}·\sqrt{8}+{\left(\sqrt{8}\right)}^2$
= $2+2·\sqrt{2}·\sqrt{2·4}+8$
= $2+2·\sqrt{2}·\sqrt{2}·\sqrt{4}+8$
= $2+2·2·2+8$
= $2+8+8$
= $18$

$\sqrt{2x}·\sqrt{8x}$
= $\sqrt{2}·\sqrt{x}·\sqrt{8}·\sqrt{x}$
= $\sqrt{2}·\sqrt{8}·\sqrt{x}·\sqrt{x}$
= $\sqrt{2}·\sqrt{2·4}·x$
= $\sqrt{2}·\sqrt{2}·\sqrt{4}·x$
= $2·2·x$
= $4x$

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

${\left(\sqrt{40}+\sqrt{10}\right)}^2$ = ${\left(\sqrt{40}\right)}^2+2·\sqrt{40}·\sqrt{10}+{\left(\sqrt{10}\right)}^2$
= $40+2·\sqrt{10·4}·\sqrt{10}+10$
= $40+2·\sqrt{4}·\sqrt{10}·\sqrt{10}+10$
= $40+2·2·10+10$
= $40+40+10$
= $90$