Hey, everyone. I need some help with dividing radicals. Here are a couple questions that I can't get the right answer to. If you can, please tell me how it's done.

1) $\displaystyle \frac{(-3\sqrt{6})(-2\sqrt{24})}{-\sqrt{18}}$

2) $\displaystyle \frac{\sqrt{a}}{1 + \sqrt{a}}$

2. 1) $\displaystyle \frac{(-3\sqrt{6})(-2\sqrt{24})}{-\sqrt{18}}$

One way is go ahead and multiply those in the numeratior,
= [6sqrt(6*24)] / [-sqrt(18)]

Then see if you can factor the new numerator so that you can cancel the denominator,
= [6sqrt(6*3*8)] / [-sqrt(18)]
= [6sqrt(8)] / [-1]
= -6sqrt(4*2)
= -6*[2sqrt(2)]

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2) $\displaystyle \frac{\sqrt{a}}{1 + \sqrt{a}}$

In cases like this, rationalize the denominator, or multiple the numerator and the denominator by the conjugate of the denominator.
The conjugate of the denominator here is [1 -sqrt(a)], so,

= [sqrt(a) *(1 -sqrt(a))] / [(1 +sqrt(a))*(1 -sqrt(a))]
= [sqrt(a) -a] / [1 -a] -------------answer.

3. I didn't quite get that because the equasions didn't show up properly.

4. $\displaystyle \frac{(-3\sqrt{6})(-2\sqrt{24})}{-\sqrt{18}} = \frac{6\sqrt{6} \cdot \sqrt{24}}{-\sqrt{6} \cdot \sqrt{3}} = -6\sqrt{\frac{24}{3}} = -6\sqrt{8} = -6\sqrt{4}\sqrt{2} = -12\sqrt{2}$

$\displaystyle \frac{\sqrt{a}}{1 + \sqrt{a}} \cdot \frac{1 - \sqrt{a}}{1 - \sqrt{a}} = \frac{\sqrt{a} - a}{1 - a}$