When multiplying radicals, you can combine them and multiply the numbers inside the radical.

$\displaystyle \sqrt{5}\cdot\sqrt{2}=\sqrt{5\cdot2}=\sqrt{10}$

$\displaystyle \sqrt{6}\cdot3\sqrt{6}$ When multiplying radicals, you can combine them and multiply the numbers inside the radical. NUmbers outside the radical are also multiplied.

$\displaystyle 3\sqrt{36}$ We can simplify this.

$\displaystyle 3\cdot6=18$

$\displaystyle \sqrt{6}\cdot\sqrt{3}$ When multiplying radicals, you can combine them and multiply the numbers inside the radical.

$\displaystyle \sqrt{18}$ We can simplify this. Lets find a perfect square.

$\displaystyle \sqrt{18}=\sqrt{9}\cdot\sqrt{2}=3\sqrt{2}$

When dividing radicals with an integer, let's simplify the radical. We need to find a perfect square. 

$\displaystyle \frac{\sqrt{24}}{4}=\frac{\sqrt{4}\cdot\sqrt{6}}{4}=\frac{2\sqrt{6}}{4}$. We can reduce.

$\displaystyle \frac{2\sqrt{6}}{4}=\frac{\sqrt{6}}{2}$

When dividing radicals with an integer, let's simplify the radical. We need to find a perfect square. 

$\displaystyle \frac{\sqrt{48}}{16}=\frac{\sqrt{16}\cdot\sqrt{3}}{16}=\frac{4\sqrt{3}}{16}$. We can reduce.

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

When dividing radicals, we multiply top and bottom by the bottom radical to ensure our denominator is an integer.

$\displaystyle \frac{\sqrt{7}}{\sqrt{5}}=\frac{\sqrt{7}\cdot\sqrt{5}}{\sqrt{5}\cdot\sqrt{5}}=\frac{\sqrt{35}}{5}$. This can't be simplified and is our answer.

When dividing radicals, we multiply top and bottom by the bottom radical to ensure our denominator is an integer.

$\displaystyle \frac{3}{\sqrt{3}}=\frac{3\cdot\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}=\frac{3\sqrt{3}}{3}=\sqrt{3}$. 

$\displaystyle \frac{2}{1+\sqrt{3}}$ To get rid of the radical in the denominator, we need to multiply top and bottom by the conjugate which is the opposite sign of the radical expression. This would be $\displaystyle 1-\sqrt{3}$.

$\displaystyle \frac{2}{1+\sqrt{3}}\cdot\frac{1-\sqrt{3}}{1-\sqrt{3}}=\frac{2-2\sqrt{3}}{1-3}=\frac{2-2\sqrt{3}}{-2}$ Remember to FOIL out when multiplying out the denominators. Now, with our answer, we can factor out a $\displaystyle -2$.

$\displaystyle \frac{2-2\sqrt{3}}{-2}=\frac{-2(-1+\sqrt{3})}{-2}=-1+\sqrt{3}$

$\displaystyle \frac{3}{1-\sqrt{2}}$ To get rid of the radical in the denominator, we need to multiply top and bottom by the conjugate which is the opposite sign of the radical expression. This would be $\displaystyle 1+\sqrt{2}$.

$\displaystyle \frac{3}{1-\sqrt{2}}\cdot\frac{1+\sqrt{2}}{1+\sqrt{2}}=\frac{3+3\sqrt{2}}{1-2}=\frac{3+3\sqrt{2}}{-1}$ Remember to use FOIL when multiplying out the denominators. Now, with out answer, we can distribute out the $\displaystyle -1$. 

$\displaystyle \frac{3+3\sqrt{2}}{-1}=-3-3\sqrt{2}$

$\displaystyle 6\sqrt{3}*2\sqrt{8}$

Multiply the outer numbers first:

$\displaystyle 12(\sqrt{3}*\sqrt{8})$

Combine radicals:

$\displaystyle 12\sqrt{24}$

Simplify the radical:

$\displaystyle 12\sqrt{6*4}$

$\displaystyle \mathbf{24\sqrt{6}}$