can someone show me the methodology for dealing with such operations of surds as these:
The question is to express as a single fraction with a rational denominator.
I would post my working out, but it has absolutely no link to the answer, and it's because I have never been taught how to solve these types of questions before. If you could explain to me the methodology for solving these and how you did it, it would be very much helpful. Thanks
$$\frac{\sqrt{2}}{\sqrt{2}-\sqrt{3}} -\frac{3}{\sqrt{2}+\sqrt{3}}$$
$$\frac{\sqrt{2}(\sqrt{2}+\sqrt{3}) - 3\sqrt{2} + 3\sqrt{3}}{(\sqrt{2}-\sqrt{3}) (\sqrt{2}+\sqrt{3})}$$
Apply $(a+b)(a-b) = a^2 - b^2$
I think you can take it from here?
To the comment, you can see this by noticing, for $b,d \not = 0$
\begin{align*} \frac{a}{b} \pm \frac{c}{d} = \frac{a}{b}\left(\frac{d}{d}\right) \pm \frac{c}{d}\left(\frac{b}{b}\right) = \frac{ ad \pm bc}{bd} \end{align*}
I think the best way is to rationalize the denominator of each term separately.
$$\frac{\sqrt2}{\sqrt2-\sqrt3}=\frac{\sqrt2(\sqrt2+\sqrt3)}{(\sqrt2-\sqrt3)(\sqrt2+\sqrt3)}=\frac{2+\sqrt6}{2-3}$$
Again, $$\frac3{\sqrt2+\sqrt3}=\frac{3(\sqrt2-\sqrt3)}{(\sqrt2+\sqrt3)(\sqrt2-\sqrt3)}=\frac{3\sqrt2-3\sqrt3}{2-3}$$
