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

$\sqrt{5}·\sqrt{\frac{1}{125}}$
= $\sqrt{5·\frac{1}{125}}$
= $\sqrt{1·\frac{1}{25}}$
= $\sqrt{\frac{1}{25}}$
= $\frac{1}{5}$

$\left(\sqrt{10}+\sqrt{90}\right)·\sqrt{10}$
= $\sqrt{10}·\sqrt{10}+\sqrt{90}·\sqrt{10}$
= $10+\sqrt{10·9}·\sqrt{10}$
= $10+\sqrt{9}·\sqrt{10}·\sqrt{10}$
= $10+3·10$
= $10+30$
= $40$

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

$\left(\sqrt{9}-7\right)·\left(\sqrt{9}+7\right)$ = ${\left(\sqrt{9}\right)}^2-7^2$
= $9-49$
= $-40$

$\sqrt{3x}·\sqrt{75x}$
= $\sqrt{3}·\sqrt{x}·\sqrt{75}·\sqrt{x}$
= $\sqrt{3}·\sqrt{75}·\sqrt{x}·\sqrt{x}$
= $\sqrt{3}·\sqrt{3·25}·x$
= $\sqrt{3}·\sqrt{3}·\sqrt{25}·x$
= $3·5·x$
= $15x$

Wir können die $\sqrt{47}$ wie Einheiten behandeln und einfach verrechnen.

$9\sqrt{47}+7\sqrt{47}$
= $16\sqrt{47}$